All categories

3 years ago

PLEASE HELP ME WITH BOTH MATH QUESTIONS ASAP!!!!!!!!!

Mathematics

2 answers:

Kazeer [188]3 years ago

8 0

Answer:

-5(c+2)=50

-5c-10=50

-5c=50+10

c=60/-5

c=-12

-2(f-10)=50

-2f+20=50

-2f=50-20

f=30/-2

f=-15

Step-by-step explanation:

Hope this helps u!!

maxonik [38]3 years ago

4 0

Answer:

the one on top= c= -12

the bottom one= f= -15

Step-by-step explanation:

You might be interested in

Which is bigger 3/4 or 2/3

tia_tia [17]

When we make it into a common denominator, find a common multiple between the denominators. For this instance, it is 3/4=9/12 2/3=8/12 When we see it with a common denominator, 3/4 is bigger than 2/3.

8 0

4 years ago

Read 2 more answers

A researcher interviews shoppers at a local mall. The mall was chosen because it was close to the researcher's house.

san4es73 [151]

Convenience sample because it was easier for the researcher to go to.

8 0

3 years ago

Read 2 more answers

Solve for x {x+4]= 6

wlad13 [49]

The solution to the equation x(x+4) = 6 is x = -2 + √10 or x = -2 - √10 after solving with the quadratic formula.

<h3>What is a quadratic equation?</h3>

Any equation of the form $\rm ax^2+bx+c=0$ where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

As we know, the formula for the roots of the quadratic equation is given by:

$\rm x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}$

We have an equation:

x(x + 4) = 6

By distributive property:

x² + 4x = 6

x² + 4x - 6 = 0

a = 1, b = 4, c = -6

Plugging all the values in the formula:

$\rm x = \dfrac{-4 \pm\sqrt{4^2-4(1)(-6)}}{2(1)}$

After calculating:

$\rm x = \dfrac{-4 \pm\sqrt{40}}{2}$

$\rm x=-2+\sqrt{10},\ \ \ or \ \ \ x=-2-\sqrt{10}$

Thus, the solution to the equation x(x+4) = 6 is x = -2 + √10 or x = -2 - √10 after solving with the quadratic formula.

Learn more about quadratic equations here:

brainly.com/question/2263981

#SPJ1

6 0

2 years ago

Read 2 more answers

Vectors...<br><br> Let u = <-3, -5>, v = <-3, 1>. Find u + v.

Anni [7]

-3 + -3 = 0
-5 + 1 = -4
vector = < 0,-4 >

3 0

3 years ago

[30 POINTS] Please help!!!

Len [333]

Answer:

Part 1) $y=1.5x+5$

Part 2) $y=-(2/3)x-(11/3)$

Part 3) $y=0.25x+2.75$

Part 4) $y=-2x+5$

Part 5) $y=0.5x-1$

Part 6) The graph in the attached figure

Step-by-step explanation:

Part 1) we have

$m=3/2=1.5$

$point(-2,2)$

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

substitute

$y-2=1.5(x+2)$

$y=1.5x+3+2$

$y=1.5x+5$

Part 2) we know that

If two lines are perpendicular

then

the product of their slopes is equal to minus one

$m1*m2=-1$

the slope of the line 1 is equal to

$m1=1.5$

Find the slope m2

$1.5*m2=-1$

$m2=-2/3$

Find the equation of the line 2

we have

$m2=-2/3$

$point(-7,1)$

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

substitute

$y-1=(-2/3)(x+7)$

$y=-(2/3)x-(14/3)+1$

$y=-(2/3)x-(11/3)$

Part 3) we have

$m=1/4=0.25$

$point(1,3)$

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

substitute

$y-3=0.25(x-1)$

$y=0.25x-0.25+3$

$y=0.25x+2.75$

Part 4) we have

$m=-2$

$b=5$ -----> y-intercept

we know that

The equation of the line into slope intercept form is equal to

$y=mx+b$

substitute the values

$y=-2x+5$

Part 5) we have that

The slope of the line 4 is equal to $-2$

the slope of the line perpendicular to the line 4 is equal to

$-2*m=-1\\m=(1/2)=0.5$

therefore

in this problem we have

$m=0.5$

$point(-2,-2)$

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

substitute

$y+2=0.5(x+2)$

$y=0.5x+1-2$

$y=0.5x-1$

Part 6)

using a graphing tool

see the attached figure

3 0

3 years ago