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3 years ago

Decimal point between -0.83 and 1.0

Mathematics

1 answer:

pantera1 [17]3 years ago

3 0

Any number between -0.83 and 1.0 ? if so then -0.63 is one number

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What does it find the GCF of each set of numbers using prime at factorizstion

lozanna [386]

You use the factor tree.
10
5 2 then you have 2 prime numbers and you have to multiply them together

3 0

2 years ago

Estimate the circumference of a circle that has a radius of 11 m simplify it.

Zanzabum

ANSWER

$C=22\pi \: m$

EXPLANATION

The circumference of a circle is calculated using the formula:

$C=2\pi \: r$

where r=11 meters is the radius of the circle.

Let us substitute the radius into the formula to obtain,

$C=2\pi \: \times 11$

This simplifies to:

$C=22\pi$

When we substitute

$\pi = 3.14$

We get

$C=22(3.14) = 69.08m$

to the nearest hundredth.

3 0

3 years ago

PLEASE I REALLY NEED HELP

Luda [366]

Answer:

<u>GUEN AND KEITH </u>

Step-by-step explanation:

both the equations are the same so you take those and do a simultaneous equation if you were to find the answer

hope this helps :)

4 0

2 years ago

Why is it that -log(x+8)=4-log(x-7) has no solution? (they are log base 2)

Lerok [7]

$-log(x+8)=4-log(x-7)$

$-\log _{10}\left(x+8\right)+\log _{10}\left(x+8\right)=4-\log _{10}\left(x-7\right)+\log _{10}\left(x+8\right)$

$0=4-\log _{10}\left(x-7\right)+\log _{10}\left(x+8\right)$

$0+\log _{10}\left(x-7\right)=4-\log _{10}\left(x-7\right)+\log _{10}\left(x+8\right)+\log _{10}\left(x-7\right)$

$\log _{10}\left(x-7\right)=\log _{10}\left(x+8\right)+4$

$\log _{10}\left(x-7\right)=\log _{10}\left(x+8\right)+\log _{10}\left(10000\right)$

$x-7=\left(x+8\right)\cdot \:10000$

$\mathrm{Solve\:}\:x-7=\left(x+8\right)10000:\quad x=-\dfrac{26669}{3333}$

$\mathrm{Verifying\:Solutions}:\quad x=-\dfrac{26669}{3333}\space\mathrm{False}$

$\mathrm{No\:Solution\:for\:x\in \mathbb{R}}$

3 0

2 years ago

Read 2 more answers

Pls help meeeee :))))

valentinak56 [21]

Answer:

31.4 inches

Step-by-step explanation:

:))

6 0

3 years ago

Read 2 more answers