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SVEN [57.7K]
3 years ago
12

What is the answer to this question ?

Mathematics
1 answer:
lesya692 [45]3 years ago
7 0
The answer to this question is (A) is the true one
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1. Which are true of x = log10 31,500? Select all that apply.
ziro4ka [17]

Given:

x=\log_{10}31,500

To find:

Select the true statements from the given options about the given value.

Solution:

We have,

x=\log_{10}31,500

It can be written as

x=\log_{10}(2^23^25^37)

x=\log_{10}(2^2)+\log_{10}(3^2)+\log_{10}(5^3)+\log_{10}(7)     [\because \log(ab)=\log a+\log b]

x=2\log_{10}2+2\log_{10}3+3\log_{10}5+\log_{10}(7)

x=2\left(0.30105\right)+2\left(0.47712\right)+3\left(0.69897\right)+0.8451

x=4.49835

Clearly, the value of x lies between 4 and 5. So, x>4 and x.

Therefore, the correct options are C and D.

4 0
2 years ago
Find the zero of polynomial 7x+5
boyakko [2]

Answer:

x=-5/7

Step-by-step explanation:

we have

f(x)=7x+5 ----> The degree of this polynomial is 1 (linear equation)

Remember that

The zero of the polynomial is the value of x when the value of f(x) is zero

so

For f(x)=0

0=7x+5

solve for x

x=-5/7 ----> the zero of the polynomial or x-intercept

4 0
3 years ago
FRREEEEEEEEEEEEEEE POINTS
kenny6666 [7]

Answer:

no no no! don't touch me there that is my no no square

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Find the general solution for cos x = 0, where n is an integer.
Romashka-Z-Leto [24]
First we solve this equation:
x = cos^-1(0) = pi/2 and 3pi/2

Now, this is the answer for angles between 0 and 2pi

but there are "negative angles" and angles larger than 2pi
so the general solution that includes all of these cases is:

x = pi/2 + n*pi

here n is any integer.
6 0
3 years ago
What is the value of (Negative 14 Superscript 0 Baseline) Superscript negative 2? Negative StartFraction 1 Over 196 EndFraction
Nookie1986 [14]

Answer:

1

Step-by-step explanation:

Assuming, we want to find the value of ((-14)^0)^{-2}

Recall that: any non-zero number exponent zero is 1.

Using this property, we simplify our expression to (1)^{-2} since (-14)^0=1

Now using the property of exponents: a^{-n}=\frac{1}{a^n}

This implies that:1^{-1}=\frac{1}{1^2}=\frac{1}{1}=1

The correct answer is 1

5 0
3 years ago
Read 2 more answers
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