Ask question

Ask question

All categories

3 years ago

13

Please help!!!!!!!!!!!!

1 answer:

kvasek [131]3 years ago

6 0

Answer:

15

Step-by-step explanation:

a

x

x

x

xd

e

3de

cr

You might be interested in

Which of the following could be the first step in solving the equation below 4^x=17

Oksanka [162]

Answer:

Choice B. log 4^x = log 17

Step-by-step explanation:

Solve 4^x=17

x is in the exponent... so the only operation to bring the x down is the Logarithm

log 4^x = log 17 take log of both sides

then x* log 4 = log 17

and then x = (log 17)/(log 4)

4 0

4 years ago

What is greater 3 over 8 or 0.75

dusya [7]

0.75 since 3/8 equals 0.375

8 0

3 years ago

Read 2 more answers

If you have 50g of sugar and want to make 25% sugar syrup,<br> how much water do you need?

Taya2010 [7]

Answer:

3(50 g) = 150 g

Step-by-step explanation:

If you mean 25% sugar by mass, then 1/4 of the total mass is sugar, and 3/4 is water. So you need 3 times as much water as sugar.

3(50 g) = 150 g

4 0

3 years ago

Two trains leave stations

Westkost [7]

Total distance 272 miles.
Total speed (when travelling toward each other) = 75+95=170 mph
Time to meet = 272/170=1.6 hours

6 0

3 years ago

Write the linearization of the function at the points indicated. (Enter your answer as an equation. Let x be the independent var

Natalija [7]

Answer:

$\displaystyle y=5+\frac{x}{10}$

$\displaystyle y=10+\frac{(x-75)}{20}$

Step-by-step explanation:

<u>Linearization</u>

It consists of finding an approximately linear function that behaves as close as possible to the original function near a specific point.

Let y=f(x) a real function and (a,f(a)) the point near which we want to find a linear approximation of f. If f'(x) exists in x=a, then the equation for the linearization of f is

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

Let's find the linearization for the function

$y=\sqrt{25+x}$

at (0,5) and (75,10)

Computing f'(x)

$\displaystyle f'(x)=\frac{1}{2\sqrt{25+x}}$

At x=0:

$\displaystyle f'(0)=\frac{1}{2\sqrt{25+0}}=\frac{1}{10}$

We find f(0)

$f(0)=\sqrt{25+0}=5$

Thus the linearization is

$\displaystyle y=f(0)+f'(0)(x-0)=5+\frac{1}{10}x$

$\displaystyle y=5+\frac{x}{10}$

Now at x=75:

$\displaystyle f'(75)=\frac{1}{2\sqrt{25+75}}=\frac{1}{20}$

We find f(75)

$f(75)=\sqrt{25+75}=10$

Thus the linearization is

$\displaystyle y=f(75)+f'(75)(x-75)=10+\frac{1}{20}(x-75)$

$\displaystyle y=10+\frac{(x-75)}{20}$

5 0

3 years ago