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GenaCL600 [577]

2 years ago

6

the table shows the number of calories jane burns while exercising. How many calories would she burn for 29 minutes. Explain. Pl

ease help me! if u don't know this then don't answer!

1 answer:

GaryK [48]2 years ago

5 0

Answer:

319

Step-by-step explanation:i just did it

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2. 2x - y = 6,<br> - x + y = -1

allochka39001 [22]

Answer:

x = 5

y = 4

Step-by-step explanation:

just substitute the value of one variable into the other

-x +y = -1

or y = -1 +x <-------- (i just added x to the right side)

now u know what y is equal to, just substitute that value into the first equation

2x - (-1+x) = 6 <-------- instead of y, i put the value of y from the 2 equation

NOW SOLVE FOR X!

2x +1 -x = 6

x + 1 = 6

x = 5

now solve for y since you have the value of x

y = -1 + x

or y = -1 + 5

y = 4

3 0

3 years ago

Read 2 more answers

Which sequence are geometric?

tiny-mole [99]

Answer:

The second one

The third one.

The last one.

Step-by-step explanation:

Hope this helps!

6 0

3 years ago

What is 7,750 to the nearest thousand

victus00 [196]

Answer:

8000

Step-by-step explanation:

7750 = 8000

thus, 7,750 to the nearest thousand is 8000

6 0

3 years ago

Read 2 more answers

Find dy/dx of the function y = √x sec*-1 (√x)

ELEN [110]

Hi there!

$\large\boxed{\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}}$

$y = \sqrt{x} * sec^{-1}(-\sqrt{x}})$

Use the chain rule and multiplication rules to solve:

g(x) * f(x) = f'(x)g(x) + g'(x)f(x)

g(f(x)) = g'(f(x)) * 'f(x))

Thus:

f(x) = √x

g(x) = sec⁻¹ (√x)

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{\sqrt{x}\sqrt{\sqrt{x}^{2} - 1}} * \frac{1}{2\sqrt{x}}$

Simplify:

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{2|x|\sqrt{{x} - 1}}$

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}$

5 0

3 years ago

Read 2 more answers

Out of 200 std in a class. 150 like math , 120 like nepali .if the number of std who like nepali only is one third of the std wh

hichkok12 [17]

<u>ANSWER: </u>

70 students like both nepali and maths.

<u>SOLUTION: </u>

Given,out of 200 students in a class. 150 like math, 120 like nepali.

The number of students who like nepali only is one third of the students who like math .

We need to find how many likes both subjects.

Let, the number of students who like only nepali be x.

Then, x is one-third of students who like maths.

x = $\frac{1}{3} \times$ students who like maths

$\begin{array}{l}{x=\frac{1}{3} \times 150} \\\\ {x=\frac{150}{3}} \\\\ {x=50}\end{array}$

Now, out of 120 students who like nepali, only 50 students like nepali alone, which means remaining students like both nepali and maths

Students who like both subjects = 120 – 50 = 70.

Hence,70 students like both nepali and maths.

3 0

3 years ago