Ask question

Ask question

All categories

3 years ago

7

Using the given points, determine Δy. (-3, -5) and (0, 10) Δy = 3 Δy = 5 Δy = 13 Δy = 15

1 answer:

juin [17]3 years ago

4 0

Answer:

Δy=(10+5)=15

Step-by-step explanation:

we know that

The formula to calculate Δy between two points is equal to

Δy=(y2-y1)

we have

(-3, -5) and (0, 10)

substitute the values

Δy=(10-(-5))=(10+5)=15

You might be interested in

Juan’s classroom is shaped like a rectangle.the room is 40 feet long and 25 feet’s wide which rectangle could be scale drawing o

Svetach [21]

I think there's an image needed to answer this question?

I tried answering without, but you do need the image.

If the length is 40 and the width is 25, the ration of length:width is 40:25.

This can be simplified to 8:5

So the answer would be any drawing with the following dimensions:

1. length 8 ft and width 5 ft

2. length 16 ft and width 10 ft

3. length 24 ft and width 15 ft

4. length 32 ft and width 20 ft

5. length 40 ft and width 25 ft

It's most likely not the last 3 because the sizes are very big, but yeah a scale drawing maintains the same ratio so I assume one of the options for your answers is gonna be one of the answers from 1 - 4 above (5 is the given dimensions so I highly doubt that's the answer)

8 0

3 years ago

Find the tangent line approximation for 10+x−−−−−√ near x=0. Do not approximate any of the values in your formula when entering

Svetllana [295]

Answer:

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Step-by-step explanation:

We are asked to find the tangent line approximation for $f(x)=\sqrt{10+x}$ near $x=0$ .

We will use linear approximation formula for a tangent line $L(x)$ of a function $f(x)$ at $x=a$ to solve our given problem.

$L(x)=f(a)+f'(a)(x-a)$

Let us find value of function at $x=0$ as:

$f(0)=\sqrt{10+x}=\sqrt{10+0}=\sqrt{10}$

Now, we will find derivative of given function as:

$f(x)=\sqrt{10+x}=(10+x)^{\frac{1}{2}}$

$f'(x)=\frac{d}{dx}((10+x)^{\frac{1}{2}})\cdot \frac{d}{dx}(10+x)$

$f'(x)=\frac{1}{2}(10+x)^{-\frac{1}{2}}\cdot 1$

$f'(x)=\frac{1}{2\sqrt{10+x}}$

Let us find derivative at $x=0$

$f'(0)=\frac{1}{2\sqrt{10+0}}=\frac{1}{2\sqrt{10}}$

Upon substituting our given values in linear approximation formula, we will get:

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}(x-0)$

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}x-0$

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Therefore, our required tangent line for approximation would be $L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$ .

8 0

3 years ago

I need help. I'm not quite sure which one to choose.

MArishka [77]

I believe it is the first one. A. $\frac{x^2}{5^2}- \frac{y^2}{5^2}\ \textgreater \ 1$

7 0

3 years ago

Read 2 more answers

Evaluate 5n for n = 2<br> Pls hurry and help- I’ll mark brainliest!!

Sveta_85 [38]

Answer:

10

Step-by-step explanation:

5 times 2 is 10.

Variables!!!!

4 0

3 years ago

A company sells fruit cups in packs of 4. The packs currently weigh 20 oz. The company plans to reduce the weight of each cup by

skad [1K]

Answer:

4 ( 5 - n )

Step-by-step explanation:

The expression:

20 – 4n

Task: Rewrite the expression for the new weight as a product of two factors.

To do this we simply have to find the highest common factor (HCF) between 20 and 4n

20 = 4 * 5

4n = 4 * n

The HCF = 4

Expressing as a product of two factors;

4 ( 5 - n )

6 0

3 years ago