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3 years ago

For a class, Mrs. Hawk brought

Mathematics

2 answers:

Mekhanik [1.2K]3 years ago

6 0

Answer:

Step-by-step explanation:

7x2=14

tangare [24]3 years ago

3 0

7C2 = 7!/[2!(7-2)!] = 7!/(2x5!) = 7x6/2 = 21
Therefore there are 21 possible combinations.

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Guys help i cant figure this out I'm stressing please help me

Mashcka [7]

Answer:

C) 0.066...

Step-by-step explanation:

1/15 = 0.066...

You could eliminate B and C as 15 (numerator from question) is not a multiple of 10. You can eliminate A as the numerator of 15 from the question is larger than 10 and will result in a number less than 0.1.

4 0

3 years ago

Verify sin^4x-sin^2x=cos^4x-cos^2x is an identity

lara [203]

ANSWER

See below

EXPLANATION

We want to verify that,

${ \sin ^{4} x} - { \sin^{2} x} = { \cos ^{4} x} - { \cos^{2} x}$

To verify this identity, we can take the left hand side simplify it to get the right hand side or vice versa.

${ \sin ^{4} x} - { \sin^{2} x} =( { \sin ^{2} x} )^{2} - { \sin^{2} x}$

${ \sin ^{4} x} - { \sin^{2} x} ={ \sin ^{2} x}({ \sin ^{2} x} - 1)$

${ \sin ^{4} x} - { \sin^{2} x} ={ \sin ^{2} x} \times - (1 - { \sin ^{2} x})$

${ \sin ^{4} x} - { \sin^{2} x} =({1 - \cos^{2} x} )\times - ({ \cos^{2} x})$

${ \sin ^{4} x} - { \sin^{2} x} =({ \cos^{2} x} - 1 )\times ({ \cos^{2} x})$

We now expand the right hand side to get:

${ \sin ^{4} x} - { \sin^{2} x} = { \cos ^{4} x} - { \cos^{2} x}$

5 0

3 years ago

Find dx/dt when y=2 and dy/dt=1, given that x^4=8y^5-240 dx/dt=

katrin2010 [14]

Answer:

The value of $\frac{dx}{dt}$ is $\frac{160}{x^3}$ .

Step-by-step explanation:

The given equation is

$x^4=8y^5-240$

We need to find the value of $\frac{dx}{dt}$ .

Differentiate with respect to t.

$4x^3\frac{dx}{dt}=8(5y^4)\frac{dy}{dt}-0$ $[\because \frac{d}{dx}x^n=nx^{n-1},\frac{d}{dx}C=0]$

$4x^3\frac{dx}{dt}=40y^4\frac{dy}{dt}$

It is given that y=2 and dy/dt=1, substitute these values in the above equation.

$4x^3\frac{dx}{dt}=40(2)^4(1)$

$4x^3\frac{dx}{dt}=40(16)(1)$

$4x^3\frac{dx}{dt}=640$

Divide both sides by 4x³.

$\frac{dx}{dt}=\frac{640}{4x^3}$

$\frac{dx}{dt}=\frac{160}{x^3}$

Therefore the value of $\frac{dx}{dt}$ is $\frac{160}{x^3}$ .

4 0

3 years ago

What is the approximate area of the shaded region under the standard normal curve below? Use the portion of the standard normal

kotegsom [21]

This region as the same area as the one in your previous question.

Pr[1 < Z < 2] = Pr[-2 < Z < -1] ≈ 0.14

5 0

3 years ago

On a map the scale is 1 inch = 125 miles. What is the actual distance between two cities of the map distance is 4 inches. ( PLS

kogti [31]

Answer:

500 miles

Step-by-step explanation:

Please mark as brainliest.

8 0

3 years ago