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3 years ago

10

Find the area of a triangle with a base (b) of 7 feet and a height (h) of 12 feet.

1 answer:

siniylev [52]3 years ago

6 0

Answer:

Step-by-step explanation:

Area = 1/2 × base × height

Base(b) = 7 feet

Height(h) = 12 feet

Area = 1/2 × 7 × 12 = 7 × 6 = 42 feet²

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| 2) through: (-5, -4), perp. to y = -x + 2

schepotkina [342]

Answer:

$y=x+1$

Step-by-step explanation:

Your generic line will be $y= mx+q$ , so you'll need two condition.

Since it has to be perpendicular to $y=(-1)x+2$ , the product of the slopes needs to be -1, or $m(-1)=-1$ , hence your "generic" becomes $y=x+q$ . Since it has to pass through [/tex](-5;-4), let's put the coordinates in there: $(-4)=1(-5)+qq-5=-4q=1$

your line becomes $y=x+1$

8 0

3 years ago

Not a real question I just need help to figure it out what is this

Rama09 [41]

Answer:

y=mx+b

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Britney lost $450 on investment which was 45% of the money she invested how much money did she invest ?

Whitepunk [10]

She would have invested $1,000 because 1,000x0.45=450. I hope that helped you!

5 0

3 years ago

Graph the equations to solve the system. <br> y=3x+3<br> 1/3y+x+1

riadik2000 [5.3K]

Given equation is y=3x+3 and $\frac{1}{3}y=x+1$

Let's simplify the 2nd equation $\frac{1}{3}y=x+1$ before we can start graph so that calculation will be easy

$\frac{1}{3}y=x+1$

multiply both sides by 3 to cancel out fractions

$3*\frac{1}{3}y=3*x+3*1$

y=3x+3

which is exactly same as the first equation so graph of both will be exactly same and solution will be infinitely many solutions.

y=3x+3 has y-intercept 3 so first point will be (0,3). Slope is 3 so rise 3 unit up then 1 right and graph the new point.

4 0

3 years ago

What is the difference?

Whitepunk [10]

Answer:

$\frac{x^2+3x -12}{(x-5) (x +3)(x+7)}$

Step-by-step explanation:

Given

$\frac{x}{x^2-2x-15} - \frac{4}{x^2+2x-35}$

Required

Calculate the difference

We start by factorizing the denominator of both fractions

$\frac{x}{x^2-2x-15} - \frac{4}{x^2+2x-35}$

$\frac{x}{x^2+3x - 5x -15} - \frac{4}{x^2+7x - 5x-35}$

$\frac{x}{x(x+3) - 5(x +3)} - \frac{4}{x(x+7) - 5(x+7)}$

$\frac{x}{(x-5) (x +3)} - \frac{4}{(x-5)(x+7)}$

Take LCM

$\frac{x(x+7) - 4(x +3)}{(x-5) (x +3)(x+7)}$

Open Brackets

$\frac{x^2+7x - 4x -12}{(x-5) (x +3)(x+7)}$

$\frac{x^2+3x -12}{(x-5) (x +3)(x+7)}$

This can't be simplified any further;

4 0

3 years ago