Ivan is saving money to buy a game. so far he has saved $30 , which is two-thirds of the total cost of the game. how much does the game cost?

(7x+3) x (4x-5)

7x*4x = 28x^2

7x*-5 = -35x

3+4x = 12x

3*-5 = -15

so you get 28x^2 -23x -15

Answer:  The required multiplied expression is [tex]28x^2-23x-15.[/tex]

Step-by-step explanation:  We are given to multiply the following linear factors and write the answer is standard form :

[tex]M=(7x+3)(4x-5)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

We will be using the following distributive property :

[tex](a+b)(c+d)=a(c+d)+b(c+d).[/tex]

From (i), we get

[tex]M\\\\=(7x+3)(4x-5)\\\\=7x(4x-5)+3(4x-5)\\\\=28x^2-35x+12x-15\\\\=28x^2-23x-15.[/tex]

Thus, the required multiplied expression is [tex]28x^2-23x-15.[/tex]

area = l x w

10 * 5 = 50

 answer is D. 50 yd^2

The correct answer is:

5 tons = 10,000 lb

10,000 lb − 570 lb = 9,430 lb = 4 tons and 1,430 lb

This answer is if you want it in tons and pounds.

Final answer:

The point on the paraboloid y = x^2 + z^2 where the tangent plane is parallel to the plane 3x + 2y + 7z = 2 is found by setting the normal vectors proportional. Solving the equations, the point is (3/4, 29/8, 7/4).

Explanation:

To find the point on the paraboloid y = x2 + z2 where the tangent plane is parallel to the plane 3x + 2y + 7z = 2, we first need to determine the normal vector of the given plane. The normal vector of the plane is defined by its coefficients, which are (3, 2, 7). For the paraboloid, we can find the normal vector at any point by taking the gradient of the function y.

The gradient of y with respect to x and z is (2x, 1, 2z). A tangent plane to the paraboloid at point (x, y, z) will have this gradient as its normal vector. To find the point where this tangent plane is parallel to the given plane, we set the gradients equal to each other up to a constant factor because parallel planes have proportional normal vectors.

Therefore, we solve the equations:

2x = 3k

1 = 2k

2z = 7k

From the second equation, k = 1/2. Substituting k into the other equations, we find x = 3/4 and z = 7/4. Now we can substitute x and z into the equation of the paraboloid to find y:

y = (3/4)2 + (7/4)2 = 9/16 + 49/16 = 58/16 = 29/8.

The point on the paraboloid where the tangent plane is parallel to the given plane is (3/4, 29/8, 7/4).

The probability that the first student picks a Snickers and then the second student also picks a Snickers is 10/91.

The question is asking for the probability that two students will each pick a Snickers from a jar containing various types of candy. We start with a total of 14 candies: 5 Snickers, 2 Butterfingers, 4 Almond Joys, and 3 Milky Ways. When the first student picks a Snickers, there are 5 possible Snickers they can pick out of the 14 total candies.

The probability for the first student is therefore  rac{5}{14}. Assuming they pick a Snickers, there are now 13 candies left in total, with 4 being Snickers. So, the second student now has a probability of  rac{4}{13} to pick a Snickers.

To find the overall probability of both events happening, we multiply the probabilities of each individual event occurring. The calculation is:

rac{5}{14} \times  rac{4}{13} =  rac{20}{182} =  rac{10}{91}

The probability that the first student picks a Snickers and then the second student also picks a Snickers is  rac{10}{91}.

The expression could be used to find the area of the isosceles triangle above is  [tex]\sqrt{40} \cdot \sqrt{40} /2[/tex]

The length of the base is the distance between the points 4+2i and 10+4i, so

Base= |10+ 4i (4+2i)| = |10+4i-4-2i|= |6 + 2i| = [tex]\sqrt{6^2 +2^2}[/tex] = [tex]\sqrt{36+4}[/tex]= √40

The middle point of the base is placed at point

4+2i+ 10 + 4i/2 =  6i +14/2 = 7+ 3i

The length of the height is the distance between the points 5+9i and 7+3i

Height = 5 +91 (7+3i)| =|5+ 9i −7 - 3i| = |−2+6i| = [tex]\sqrt{(-2)^2 + 6^2} = \sqrt {4+36}[/tex] = √40

So, the area of the triangle is

[tex]A= 1/2 \cdot Base \cdot Height= \sqrt{40} \cdot \sqrt{40} /2[/tex]

Therefore, The expression could be used to find the area of the isosceles triangle above is  [tex]\sqrt{40} \cdot \sqrt{40} /2[/tex]

The probable question may be:

An isosceles triangle’s altitude will bisect its base. Which expression could be used to find the area of the isosceles triangle above?

Points on the graph of the triangle are (5+9i), (10+4i), and  (4+2i).

A. \sqrt{40} \cdot \sqrt{40} /2

B. \sqrt{40} \cdot \sqrt{68} /2

C. \sqrt{232} \cdot \sqrt{288} /2

D.  \sqrt{232} \cdot \sqrt{164} /2

Final answer:

To find the area of an isosceles triangle when given the length of the base and the altitude, you can use the formula A = ½ × base × height. The altitude of an isosceles triangle will bisect its base, so you can divide the base in half to find the length of the base above the altitude.

Explanation:

An isosceles triangle's altitude will bisect its base. To find the area of the isosceles triangle, we can use the formula A = ½ × base × height. Since the altitude bisects the base, we can divide the base in half to find the length of the base above the altitude. Let's say the length of the base is 2x and the length of the altitude is h. So, the expression to find the area of the isosceles triangle is A = ½ × (2x) × h = xh.

The equation of the line in slope-intercept form would be; y=1/2x+5.

The slope is the ratio of the vertical changes to the horizontal changes between two points of the line.

(a) The slope of line passing through two points is given by;

[tex]m = \dfrac{y-q}{x-p}[/tex]

Here m = 4 - 9/ -2 - 8

m = 5/10

m = 1/2

(b) The equation of line passing through a point and having slope is given by ;

y-y₁ = m(x-x₁)

Here, y-4 = 1/2(x+2)

(c) The slope intercept form;

y-4=1/2(x+2)

y-4=1/2x+1

y=1/2x+5

Hence, the equation of the line in slope-intercept form would be; y=1/2x+5.

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Answer with explanation:

1. The given rational expression is

   [tex]\rightarrow\frac{x^2+5 x+6}{x^2-9}\\\\\rightarrow\frac{x^2+5 x+6}{(x-3)(x+3)}[/tex]

The function is not defined ,when

 →(x-3)(x+3)=0

→x-3≠0 ∧ x+3≠0

→x≠3, ∧ x ≠ -3

⇒Option C and D

→x≠−3

→x≠3

2.

[tex]\rightarrow\frac{4 x}{x-3}+\frac{2}{x^2-9}=\frac{1}{x+3}\\\\\rightarrow\frac{4 x}{x-3}+\frac{2}{(x-3)(x+3)}=\frac{1}{x+3}\\\\\rightarrow\frac{4 x(x+3)+2}{(x-3)(x+3)}=\frac{1}{x+3}\\\\\rightarrow4 x(x+3)+2=\frac{(x-3)(x+3)}{x+3}\\\\\rightarrow4x^2+12 x+2=x-3\\\\\rightarrow4x^2+11x+5=0\\\\ \text{Using Discriminant method for a quadratic equation}\\\\ax^2+bx +c=0\\\\x=\frac{-b\pm\sqrt{D}}{2 a}\\\\D=b^2-4 ac\\\\x=\frac{-11\pm\sqrt{121-80}}{2 \times 4}\\\\x=\frac{-11\pm\sqrt{41}}{8}[/tex]

None of the option

3.

[tex]\rightarrow \frac{1}{x}+\frac{1}{x-3}=\frac{x-2}{x-3}\\\\\rightarrow\frac{x-3+x}{x(x-3)}=\frac{x-2}{x-3}\\\\\rightarrow2x-3=\frac{x(x-3)(x-2)}{x-3}\\\\\rightarrow 2 x-3=x^2-2 x\\\\\rightarrow x^2-4x+3=0\\\\\rightarrow (x-1)(x-3)=0\\\\x=1,3[/tex]

For, x=3 , the equation is not defined.

So, there is single solution which is , x=1.

Option B:→ There is only one solution: x = 1.

The solution x = 3 is an extraneous solution.

4.

[tex]\rightarrow \sqrt{3}x+1-x+3=0\\\\\rightarrow \sqrt{3}x -x=-4\\\\\rightarrow x(\sqrt{3}-1)=-4\\\\\rightarrow x=\frac{-4}{\sqrt{3}-1}\\\\\rightarrow x=\frac{-4\times(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\\\\x=\frac{-4\times(\sqrt{3}+1)}{2}\\\\x=-2(\sqrt{3}+1)[/tex]

None of the option

1) Options c) and d) are correct.

2) None of the options are correct.

3) Option B) is correct.

4) None of the options are correct.

Step-by-step explanation:

1) Given :   [tex]4x^2+12x+2=x-3[/tex]

  Expression --         [tex]\dfrac{x^2+5x+6}{x^2-9}[/tex]

  Solution :

  [tex]\dfrac{x^2+5x+6}{x^2-9}=\dfrac{x^2+3x+2x+6}{(x+3)(x-3)}[/tex]

  [tex]\dfrac{x^2+5x+6}{x^2-9}=\dfrac{(x+3)(x+2)}{(x+3)(x-3)}[/tex]

  Therefore, [tex]\rm x\neq 3 \; and\;x\neq -3[/tex] ,option c) and d) is correct.

2) Given :

    Expression - [tex]\dfrac{4x}{x-3}+\dfrac{2}{x^2-9}=\dfrac{1}{x+3}[/tex]

    Solution :

     [tex]\dfrac{4x}{(x-3)}+\dfrac{2}{(x+3)(x-3)}=\dfrac{1}{(x+3)}[/tex]

     [tex]\dfrac{4x(x+3)+2}{(x-3)(x+3)}=\dfrac{1}{(x+3)}[/tex]

     [tex]4x^2+12x+2=x-3[/tex]

     [tex]4x^2+11x+5=0[/tex]

     [tex]x=\dfrac{-11\pm\sqrt{121-80} }{8}[/tex]

     [tex]x = \dfrac{-11\pm\sqrt{41} }{8}[/tex]

    None of the options are correct.

3) Given :

  Expression -     [tex]\dfrac{1}{x}+\dfrac{1}{x-3}=\dfrac{x-2}{x-3}[/tex]     ----- (1)

   Solution :

   [tex]\dfrac{x-3+x}{(x)(x-3)}=\dfrac{x-2}{x-3}[/tex]

   [tex]2x-3=x(x-2)[/tex]

    [tex]x^2-4x+3=0[/tex]

    [tex]x^2-3x-x+3=0[/tex]

    [tex](x-3)(x-1)=0[/tex]

At x = 3 equation (1) is not define. Therefore, the correct answer is option

B) There is only one solution: x = 1. The solution x = 3 is an extraneous solution.

4) Given :

   Exprression -  [tex](\sqrt{3}x +1)-x+3=0[/tex]

   Solution :

   [tex]x(\sqrt{3}-1 )=-4[/tex]

   [tex]x=\dfrac{-4}{\sqrt{3}-1 }[/tex]

   None of the options are correct.

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The value of [tex]\int\limits^._R {xy^2} \, dA[/tex]  is 16.

The calculation of an integral is called integration. In math, many useful quantities like areas, volumes, displacement, and so on can be found using integrals. When we discuss integrals, we typically refer to definite integrals. Antiderivatives make use of the indefinite integrals. Apart from differentiation, one of the two major calculus topics in mathematics is integration.

Given xy =3, xy = 7,

xy² = 3, xy² = 7

s = xy and t = xy²

dividing t by s we get

t/s = xy²/xy

y = t/s

and x = s/y = s²/t

now differentiate x and y partially with respect to s and t

∂x/∂s = 2s/t

∂x/∂t = -s²/t²

∂y/∂s = -t/s²

∂y/∂t = 1/s

The Jacobian is

∂(x, y)/∂(s, t) = [tex]\left|\begin{array}{ccc}dx/ds&dx/dt\\dy/ds&dy/dt\end{array}\right|[/tex]

∂(x, y)/∂(s, t) =[tex]\left|\begin{array}{ccc}2s/t&-s^{2}/t^{2} \\-t/s^{2} &1/s\end{array}\right|[/tex]

∂(x, y)/∂(s, t) =2/t - 1/t

∂(x, y)/∂(s, t) = 1/t

so for  [tex]\int\limits^._R {xy^2} \, dA[/tex] = [tex]\int\limits^a_b \int\limits^a_b {t}\frac{d(x, y)}{d(s, t) } \, dsdt[/tex]

for (s, t) (3 ≤ s ≤7, 3 ≤ t  ≤ 7)

= [tex]\int\limits^7_3 \int\limits^7_3 {t}*1/t } \, dsdt[/tex]

= [tex]\int\limits^7_3ds \int\limits^7_3 dt[/tex]

=[s]₃⁷ [t]₃⁷

= (7 - 3)(7 - 3)

= 16

Hence the value is 16.

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Answer:

Notice that the given line is vertical, which means its slope is undetermined.

We can demonstrate that using the slope definition:

[tex]m=\frac{y_{2} -y_{1} }{x_{2}-x_{1} }[/tex]

[tex]m=\frac{1-0}{2-2}=\frac{1}{0}=IND[/tex]

Which is undetermined because null denominators cannot be determined.

Now, the Point-Slope form of a line refers to expressin the equation with one point and its slope. So, if we cannot determine the slope, we cannot determine the point-slope form neither. In other words, we need to have a determined slope to express it in this form.

Answer:

option B

Step-by-step explanation:

A. [tex]3x^2 - 11x + 4 = 0[/tex]

3*4 = 12

We find out two factors whose sum is -11 and product is 12

1 times 12 = 12

To get sum -11 , then one factor should be negative. So, factoring is not possible.

B .  [tex]3x^2 - 11x - 4 = 0[/tex]

3*(-4) = -12

We find out two factors whose sum is -11 and product is -12

1 times (-12) = -12

1 + (-12) = -11

So two factors are 1  and -12

Split the middle term -11x using factors 1  and -12

So equation becomes

[tex]3x^2 + 1x - 12x - 4 = 0[/tex]

Now group first two terms and last two terms

[tex](3x^2 + 1x)+ (- 12x - 4) = 0[/tex]

[tex]x(3x+ 1)-4(3x +1)=0[/tex]

(3x+1)(x-4)=0

Now we set each parenthesis =0  and solve for x

3x+1 =0  , subtract 1 on both sides

3x = -1 ( divide both sides by 3)

x= -1/3

Now we set x-4=0

add 4 on both sides

so x=4

Option B is correct