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3 years ago

What's the radical of 40

Mathematics

1 answer:

garik1379 [7]3 years ago

7 0

2.............................

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How do you multiply a whole number with a mixed number

marta [7]

You would change the whole number to a fraction by putting it over 1 and make the mixed number an improper fraction by multiplying the whole number part of the mixed number by the denominator and then and the numerator

6 0

3 years ago

Which is greater 15/27 or 16/24

Artemon [7]

Answer:

16/24

Step-by-step explanation:

let’s use the GCF of both which is 3 for 15/27

15/27 divided by 3 is 5/9

for 16/24 the gcf is 8

16/24 / 8 is 2/3

now we have to multiply 2/3 times 3

2/3 times 3 6/3 which is 2

so 16/24 is greater.

3 0

3 years ago

An observer (O) is located 900 feet from a building (B). The observer notices a helicopter (H) flying at a 49° angle of elevatio

Firdavs [7]

Answer:

1035.33 feet

Step-by-step explanation:

h=900tan49

=1035.33 feet

I hope it helps

6 0

3 years ago

Four gallons of gasoline weigh 25 pounds . find the unit rate in pounds per gallon.

notka56 [123]

There are several information's that are already given in the question. Based on those information's the answer to the question can be easily deduced.
Weight of 4 gallons of gasoline = 25 pounds
Then
Weight of 1 gallon of gasoline = 25/4 pounds
= 6.25 pounds
From the above deduction, we can easily conclude that the unit rate in pounds per gallon of gasoline is 6.25 pounds/gallon. I hope the answer helps you.

7 0

3 years ago

Read 2 more answers

Identify the "inside function" u = f(x) and the "outside function" y = g(u). Then find dy/dx using the Chain Rule.

skad [1K]

DfLet $f(x)=\sec x$ and $g(x)=\sqrt x$ . Then

$y=\sec\sqrt x=\sec(g(x))=f(g(x))=f\circ g(x)$

By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm du}\dfrac{\mathrm du}{\mathrm dx}$

where $u=g(x)=\sqrt x$ , so that $y=f(g(x))=f(u)=\sec u$ . We have

$\dfrac{\mathrm du}{\mathrm dx}=\dfrac1{2\sqrt x}$
$\dfrac{\mathrm d\sec u}{\mathrm du}=\sec u\tan u$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\sec u\tan u\dfrac{\mathrm dy}{\mathrm du}=\dfrac{\sec\sqrt x\,\tan\sqrt x}{2\sqrt x}$

5 0

3 years ago