Ask question

Ask question

All categories

igor_vitrenko [27]

3 years ago

11

HELP ME PLEASE!!!!! IM DYING WITH ALL THIS MATH HW

2 answers:

sashaice [31]3 years ago

7 0

Nxjdkdnc knockzjznsbdbxbzmzmmd xbxidnrnfu. Xizneshzudnd check x

trasher [3.6K]3 years ago

5 0

3x=129 , x =43 distribute 3 to get x by itself

You might be interested in

Let f(x) = 7x^2-5x+3 and g(x) = 2x^2+4x-6

Troyanec [42]

Answer:

$\large\boxed{A.\ f(x)+g(x)=9x^2-x-3}\\\boxed{B.\ f(x)-g(x)=5x^2-9x+9}\\\boxed{g(x)-f(x)=-5x^2+9x-9}$

Step-by-step explanation:

$f(x)=7x^2-5x+3,\ g(x)=2x^2+4x-6\\\\A:\\f(x)+g(x)=(7x^2-5x+3)+(2x^2+4x-6)\\f(x)+g(x)=7x^2-5x+3+2x^2+4x-6\qquad\text{combine like terms}\\f(x)+g(x)=(7x^2+2x^2)+(-5x+4x)+(3-6)\\f(x)+g(x)=9x^2-x-3$

$B:\\f(x)-g(x)=(7x^2-5x+3)-(2x^2+4x-6)\\f(x)+g(x)=7x^2-5x+3-2x^2-4x+6\qquad\text{combine like terms}\\f(x)-g(x)=(7x^2-2x^2)+(-5x-4x)+(3+6)\\f(x)+g(x)=5x^2-9x+9$

$C:\\g(x)-f(x)=(2x^2+4x-6)-(7x^2-5x+3)\\g(x)-f(x)=2x^2+4x-6-7x^2+5x-3\qquad\text{combine like terms}\\g(x)-f(x)=(2x^2-7x^2)+(4x+5x)+(-6-3)\\g(x)-f(x)=-5x^2+9x-9$

3 0

3 years ago

Line F has a slope of. -6/3 and line G has a slope of -8/4.

Darina [25.2K]

Given:

Slope of line F = $-\dfrac{6}{3}$

Slope of line G = $-\dfrac{8}{4}$

To find:

The conclusion about distinct lines F and G.

Solution:

We have,

Slope of line F = $-\dfrac{6}{3}$

= $-2$

Slope of line G = $-\dfrac{8}{4}$

= $-2$

The slopes of lines F and G are equal and we know that the slopes of two parallel lines are always equal.

Therefore, the line F and line G are parallel to each other.

7 0

3 years ago

Can someone help me with this question???

JulsSmile [24]

Neither train a or train b

5 0

3 years ago

Read 2 more answers

What is the volume of the following triangular prism?

Greeley [361]

Answer:

Option C. $42\ yd^{3}$

Step-by-step explanation:

we know that

The volume of the triangular prism is equal to

$V=BH$

where

B is the area of the triangular base

H is the height of the prism

Convert 9 ft to yd

Remember that

1 ft=1/3 yd

9 ft=9*(1/3)=3 yd

<em>Find the area of the base B</em>

$B=\frac{1}{2}(3)(4)=6\ yd^{2}$

we have

$H=7\ yd$

substitute the values in the formula

$V=BH$

$V=(6)(7)=42\ yd^{3}$

5 0

3 years ago

Read 2 more answers

How many points of intersection are there between the graphs of the hyperbola and ellipse?

N76 [4]

The point of intersection is the point where two or more functions meet.

The graphs of the hyperbola and ellipse have 0 point of intersection

The given parameters are:

$\mathbf{3x^2 - 4y^2 + 4x - 8y + 4 = 0}$

$\mathbf{3x^2 + y^2 + 4x - 3y + 4 = 0}$

To determine the points of intersection, we simply equate both equations.

So, we have:

$\mathbf{3x^2 + y^2 + 4x - 3y + 4 = 3x^2 - 4y^2 + 4x - 8y + 4 }$

Cancel out the common terms

$\mathbf{y^2 - 3y = - 4y^2 - 8y }$

Collect like terms

$\mathbf{y^2 +4y^2= 3y - 8y }$

$\mathbf{5y^2= - 5y }$

Divide through by 5

$\mathbf{y^2= - y}$

Divide through by y

$\mathbf{y= -1}$

<em>The system of equation does not give room to calculate the x-coordinate at the x-axis.</em>

<em />

Hence, the graphs of the hyperbola and ellipse have <em>0 point of intersection</em>

Read more about points of intersection at:

brainly.com/question/13373561

4 0

2 years ago