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3 years ago

5

write a real-world problem that involves a positive and negative value. Then graph the values used in the problem on a number li

ne. Compare the values.

1 answer:

vovikov84 [41]3 years ago

7 0

Jeremiah had 200 cat treats and 200 dogs treats and he gave Makayla 120 cat treats and 168 treats all together. How many treats does Jeremiah have left? And how many dog treats does does Makayla have ?

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Math help soon please

mixas84 [53]

The slope is 3/5 just count the units up and to the side

5 0

3 years ago

First the store raised the price of a pot by 20%. Then they announced a 20% discount on the pot. Is the customer going to pay mo

andrezito [222]

Answer:

less

Step-by-step explanation:

An increase of 20% makes the price 120% of the original price.

A discount of 20% makes the discounted price 80% of the previous price.

Original price: x

20% increase in price: 1.2 * x

20% discount over the raised price: 0.8 * 1.2 * x

Now we multiply the numbers in the last expression and compare the result with x.

0.8 * 1.2 * x = 0.96x

0.96 is less than x, so the customer pays less than the original price.

7 0

2 years ago

40/6 as a mixed number and in simpliest form

givi [52]

Answer is : 6 4/6
in simplest form is : 6 2/3

8 0

3 years ago

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

so

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

What is the slope of o line that passes through (-4,-14) and (19,11)?

DerKrebs [107]

Answer:

D, see description below.

There is a possible typo in the problem since these points have slope 25/23.

Step-by-step explanation:

To find the slope of a line, calculate the distance between points with a ratio of vertical to horizontal distance. This is done using the formula:

$m=\frac{y_2-y_1}{x_2-x_1}$

Here the points are (-4,-14) and (19,11). Substitute and simplify.

$m=\frac{11--14}{19--4}=\frac{11+14}{19+4}=\frac{25}{23}$

I think there is a typo in your points (-4,-14) and (19,11) because the answer doesn't align to the answer choices. But based on them, I would answer D or 24/23 since it's the closest value.

6 0

3 years ago