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3 years ago

What adds to negative 4 and multiplys to negative 16

Mathematics

1 answer:

TEA [102]3 years ago

4 0

12 bdbdjfjcifnfskkHxhxudhdhxjjzjzkzjxhfhd idk

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Graph A line goes through the point (6,–2) and has a slope of –3. What is the value of a if the point (a,7) lies on the line?

3241004551 [841]

Answer:

The value of a is 3

Step-by-step explanation:

First you have to make an equation based on the point and slope that you were given. The formula for this is y-(y cordinate of point)=slope(x-x cordinate of slope)

y--2=-3(x-6), which simplifies to y+2=-3(x-6)

In (a,7), 7 is the y coordinate. We have to plug 7 in for the y in our equation.

7+2=-3(x-6),

Combine like terms on the left side to get to 9=-3(x-6)

Multiply out on the left to get 9=-3x+18

Subtract 18 from the left side to get -9=-3x

Divide both sides by negative 3 to get 3=x (remember x is a)

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3 years ago

After unloading 96 pounds of mulch from the truck, there were still 24 pounds

Dennis_Churaev [7]

Answer:

120

Step-by-step explanation:

To find the answer you just have to add 24 and 96 together to get 120.

4 0

3 years ago

If h(x)=1-2/3x, find h(-6).

Alex

Answer:

Step-by-step explanation:

Put -6 where x is and do the arithmetic.

$h(x)=1-\dfrac{2}{3}x\\\\h(-6)=1-\dfrac{2}{3}(-6)=1-\dfrac{-12}{3}=1-(-4)=1+4\\\\\boxed{h(-6)=5}$

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3 years ago

How can I be smart wear I don't need to cheat to get high grades

ss7ja [257]

try your best

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Fine length of BC on the following photo.

MrMuchimi

Answer:

$BC=4\sqrt{5}\ units$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

In the right triangle ACD

Find the length side AC

Applying the Pythagorean Theorem

$AC^2=AD^2+DC^2$

substitute the given values

$AC^2=16^2+8^2$

$AC^2=320$

$AC=\sqrt{320}\ units$

simplify

$AC=8\sqrt{5}\ units$

step 2

In the right triangle ACD

Find the cosine of angle CAD

$cos(\angle CAD)=\frac{AD}{AC}$

substitute the given values

$cos(\angle CAD)=\frac{16}{8\sqrt{5}}$

$cos(\angle CAD)=\frac{2}{\sqrt{5}}$ ----> equation A

step 3

In the right triangle ABC

Find the cosine of angle BAC

$cos(\angle BAC)=\frac{AC}{AB}$

substitute the given values

$cos(\angle BAC)=\frac{8\sqrt{5}}{16+x}$ ----> equation B

step 4

Find the value of x

In this problem

$\angle CAD=\angle BAC$ ----> is the same angle

equate equation A and equation B

$\frac{8\sqrt{5}}{16+x}=\frac{2}{\sqrt{5}}$

solve for x

Multiply in cross

$(8\sqrt{5})(\sqrt{5})=(16+x)(2)\\\\40=32+2x\\\\2x=40-32\\\\2x=8\\\\x=4\ units$

$DB=4\ units$

step 5

Find the length of BC

In the right triangle BCD

Applying the Pythagorean Theorem

$BC^2=DC^2+DB^2$

substitute the given values

$BC^2=8^2+4^2$

$BC^2=80$

$BC=\sqrt{80}\ units$

simplify

$BC=4\sqrt{5}\ units$

7 0

3 years ago