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OleMash [197]
3 years ago
6

Please help me find the remainders

Mathematics
1 answer:
Bingel [31]3 years ago
4 0

Answer:

- 2 and 7

Step-by-step explanation:

A polynomial P(x) divided by (x + h) has remainder f(- h) ← remainder theorem

P(x) divided by (x + 5) has remainder P(- 5) = - 2 ( from result given )

P(x) divided by (x - 3) has remainder P(3) = 7 ( from result given )

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What is a circumference of a smaller circle the raduis of a smaller circle 3inch and the the raduis of a larger is circle is 7so
Art [367]

Answer:

18.85 inches.

Step-by-step explanation:

We have been given that the radius of a smaller circle 3 inch. We are asked to find the circumference of smaller circle.

We know that circumference of a circle is equal to 2\pi r, where r represents radius of circle.

\text{Circumference of smaller circle}=2\pi (3\text{ inches})

\text{Circumference of smaller circle}=6\pi\text{ inches}

\text{Circumference of smaller circle}=18.84955592\text{ inches}

\text{Circumference of smaller circle}\approx 18.85\text{ inches}

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8 0
3 years ago
3x + 15 *
Temka [501]
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7 0
3 years ago
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What is the image of (6,−3) after a dilation by a scale factor of 1/3 centered at the origin?
attashe74 [19]

Answer:

(2,-1)

Step-by-step explanation:

Multiply everything by 1/3:

6*1/3=2

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6 0
2 years ago
Find the derivative.
Aleksandr [31]

Answer:

Using either method, we obtain:  t^\frac{3}{8}

Step-by-step explanation:

a) By evaluating the integral:

 \frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du

The integral itself can be evaluated by writing the root and exponent of the variable u as:   \sqrt[8]{u^3} =u^{\frac{3}{8}

Then, an antiderivative of this is: \frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}

which evaluated between the limits of integration gives:

\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}

and now the derivative of this expression with respect to "t" is:

\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

g(x)=\int\limits^x_a {f(t)} \, dt

is continuous on [a,b], differentiable on (a,b) and  g'(x)=f(x)

Since this this function u^{\frac{3}{8} is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}

5 0
3 years ago
Help me plz it’s due today
Serggg [28]

Answer:

A

Step-by-step explanation: Your salary will be greater than or equal to $46,000.

3 0
3 years ago
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