All categories

3 years ago

$7 is what percent of $10?

Mathematics

2 answers:

malfutka [58]3 years ago

4 0

Its 70%. hope it helped

Phantasy [73]3 years ago

4 0

So the keyword is "of" so therefore we multiply 7 x 10

Then we would get 70$

You might be interested in

A board 60 in. long is cut into two parts so that the longer piece is 5 times the shorter. What are the lengths of the two piece

EastWind [94]

Answer:

Long is 50 in

short is 10 in

Step-by-step explanation:

60=6x

x=10

Long is 50 in

short is 10 in

8 0

3 years ago

X^2+x-5=0nsolve using the quadratic formula

Dennis_Churaev [7]

Answer:

57776esae r esspustae

Step-by-step explanation:

8 0

3 years ago

A test car went 64 miles on 2 gallons in the morning, 16 miles on 1/2 gallon at noon, and 32 miles on 1 gallon at 5 PM.Is the re

kolbaska11 [484]

It is proportional because the relationship between gallons of gas and distance is constant and linear. The rate is 32 miles per gallon. You can also call 32 the slope of this linear relationship.

8 0

3 years ago

Brian needs $315 to buy a new cell phone. He has already saved $115 and plans to save an additional

Juliette [100K]

Answer:

34 weeks

Step-by-step explanation:

Hello!

We can set up an equation to solve for the number of weeks.

Let w = the number of weeks he is saving up.

The equation is $315 = 115 + 10w - 140$

315 is the target amount
115 is the current amount
10w is the rate of growth
140 is the current debt

<h3>Solve for w</h3>

$315 = 115 + 10w - 140$
$315 = -25 + 10w$
$340 = 10w$
$340/10 = w$
$34 = w$

He needs to save up for 34 weeks.

5 0

2 years ago

Use calculus to find the absolute maximum and minimum values of the function. (round all answers to three decimal places.) f(x)

Allisa [31]

Part A:

Given the function $f(x)=x+2\cos(x)$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow1-2\sin{x}=0 \\ \\ \Rightarrow2\sin{x}=1 \\ \\ \Rightarrow\sin{x}= \frac{1}{2} \\ \\ \Rightarrow x=\sin^{-1}{\frac{1}{2}}= \frac{\pi}{6}$

Using the second derivative test,

$f''(x)=-2cosx \\ \\ \Rightarrow f''\left( \frac{\pi}{6} \right)=-2\cos{\left( \frac{\pi}{6} \right)}=-1.732$

Since the second derivative gives a negative number, the given function has a maximum point at $x=\frac{\pi}{6}$ .

And the maximum point is given by:

$f\left( \frac{\pi}{6} \right)=\frac{\pi}{6}+2\cos\left( \frac{\pi}{6} \right) \\ \\ =0.5236+2(0.8660)=0.5236+1.732 \\ \\ =\bold{2.256}$

i.e. $\left(\frac{\pi}{6},\ 2.256\right)$

Part B:

Given the function $f(x)=e^{-x}-e^{-2x}$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow-e^{-x}+2e^{-2x}=0 \\ \\ \Rightarrow2e^{-2x}=e^{-x} \\ \\ \Rightarrow2e^{-x}=1 \\ \\ \Rightarrow e^{-x}=\frac{1}{2} \\ \\ \Rightarrow-x=\ln \frac{1}{2}=-0.6931 \\ \\ \Rightarrow x=0.6931$

Using the second derivative test,

$f''(x)=e^{-x}-4e^{-2x} \\ \\ \Rightarrow f''(0.6931)=e^{-0.6931}-4e^{-2(0.6931)} \\ \\ =0.5-4e^{-1.386}=0.5-4(0.25)=0.5-1 \\ \\ =-0.5$

Since the second derivative gives a negative number, the given function has a maximum point at $x=0.6931$ .

And the maximum point is given by:

$f(0.6931)=e^{-0.6931}-e^{-2(0.6931)} \\ \\ =0.5-e^{-1.386}=0.5-0.25=\bold{0.25}$

i.e. (0.693, 0.25)

3 0

3 years ago