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Nikitich [7]
3 years ago
14

What’s the answer ? Help this is a test

Mathematics
1 answer:
natali 33 [55]3 years ago
7 0

Answer:

It is D.

Step-by-step explanation:

I hope this helps.

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Show that the equation 2x + 3 cos x + e ^ x = 0 has a root on the interval [- 1, 0]
ExtremeBDS [4]

If <em>x</em> = -1, you have

2(-1) + 3 cos(-1) + <em>e</em> ⁻¹ ≈ -0.0112136 < 0

and if <em>x</em> = 0, you have

2(0) + 3 cos(0) + <em>e</em> ⁰ = 4 > 0

The function <em>f(x)</em> = 2<em>x</em> + 3 cos(<em>x</em>) + <em>eˣ</em> is continuous over the real numbers, so the intermediate value theorem applies, and it says that there is some -1 < <em>c</em> < 0 such that <em>f(c)</em> = 0.

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3 years ago
Numbers with the same absolute value are__________
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They are either the same, or opposite number.
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3 years ago
Read 2 more answers
Mary has 15 points with the total value of $1.75 if the coins are nickels and quarters how many of each kind are there
torisob [31]

Answer:

The number of nickel coins is 10 and the number of quarter coins is 5

Step-by-step explanation:

<u><em>The correct question is</em></u>

Mary has 15 coins with the total value of $1.75 if the coins are nickels and quarters how many of each kind are there

Let

x ----> the number of nickel coins

y ----> the number of quarter coins

Remember that

1\ nickel=\$0.05

1\ quarter=\$0.25

we know that

Mary has 15 coins

so

x+y=15 -----> equation A

The total value of the coins is $1.75

so

0.05x+0.25y=1.75 ----> equation B

Solve the system by graphing

Remember that the solution is the intersection point both graphs

using a graphing tool

The solution is the point (10,5)

therefore

The number of nickel coins is 10 and the number of quarter coins is 5

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3 years ago
PLEASE HELP WILL MARK BRAINLIEST!!!!!
trapecia [35]

The numbers inside the parenthesis is the value of x, uses the equation associated with the value of x and solve.

H(1) Since 1 fits 1≤x≤3 use x^3

x^3 1^3 = 1

h(1) = 1

h(4) 4 fits x >3, so use 5

h(4) = 5

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kap26 [50]
It's 80.75 for the answer!
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3 years ago
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