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3 years ago

Find the slope of ther line that goes through the points (-9,-4) and (-3,-6)

Mathematics

1 answer:

sleet_krkn [62]3 years ago

5 0

Answer:

slope = - $\frac{1}{3}$

Step-by-step explanation:

Calculate the slope m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (- 9, - 4) and (x₂, y₂ ) = (- 3, - 6)

m = $\frac{-6+4}{-3+9}$ = $\frac{-2}{6}$ = - $\frac{1}{3}$

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Rzqust [24]

Answer:

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2 years ago

What is 1+ 25372 +62719+ 82837?

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2 years ago

Read 2 more answers

3. i) Determine the volume of the cylinder above to the nearest cubic centimetre. (3 points)

klemol [59]

Answer:

1480cm³

Step-by-step explanation:

The Equation for the Volume of a Cylinder is V=Bh, where B is the area of the circle (A= πr²).

We are given all of the things we need, we just have to plug them into the equation.

First let's find B using the radius(r) which is given to be 6.4cm, so B=πr²

B=π(6.4cm)²

Plug that into a calculator to get 128.6796351cm²

Now we can plug that into the original equation, V=Bh.

V=(128.6796351cm²)h

h is also given to be 11.5cm, so lets plug that in.

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3 0

3 years ago

Lydia's mom earns $4 in rewards for every $75 she spends at the grocery store. If she earned $44 in rewards last month, how much

8_murik_8 [283]

Answer:

$495

Step-by-step explanation:

Take baseline amount of 75 dollars and divide it by 4. This will get you to the answer 11.25. Use 11.25 and multiply it by what she received in rewards. 11.25 times 44 equals....495!

6 0

2 years ago

One morning, John drive 5 hours before stopping to eat. After lunch, he. increased his speed by 10 mph. If he completed a 500-mi

Marrrta [24]

So he drove 5hrs, before lunch, now, he was going at speed, say "r", after he ate, he sped up by 10mph, so, his rate is whatever "r" is, plus 10, or " r + 10 ", after lunch

we know the whole trip took 10hrs, so, is 5hrs before lunch and 5hrs after lunch

we also know the whole trip was 500miles, so if he drove, say "d" miles before lunch, after lunch he drove the slack after lunch, or " 200 - d "

recall, your d = rt, distance = rate * time

$\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{before lunch}&d&r&5\\
\textit{after lunch}&500-d&r+10&5
\end{array}
\\\\\\

\begin{cases}
\boxed{d}=5r\\
500-d=(r+10)5\\
----------\\
500-\boxed{5r}=(r+10)5
\end{cases}
\\\\\\
\cfrac{500-5r}{5}=r+10\implies 100-r=r+10$

and I'm sure you know what that is

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2 years ago