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3 years ago

Solve for x. 12x - 39 <_9 AND -4x + 3 <-6

Mathematics

1 answer:

zlopas [31]3 years ago

5 0

Left:
Add 39 to both sides
Leaves you with 12x ≤ 48
Divide both sides by 12
Answer: x ≤ 4

Right question:
Subtract 3 from both sides
Leaves you with: 4x < -9
Divide both sides by -4
Answer: x > 9/4

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16 because it is even I believe.

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Given: △ACM, m∠C=90°, CP ⊥ AM . AC:CM=3:4, MP-AP=1. Find AM.

lana [24]

Answer:

$AM=\frac{25}{7}$

Step-by-step explanation:

It is given that AC:CM = 3:4.

Therefore, let AC = 3k and CM = 4k

It is given that Δ ACM is right angled.

Therefore,

$AM^{2} =AC^{2} +CM^{2}$

$=(3k)^{2} +(4k)^{2}$

$=9k^{2} +16k^{2}$

$=25k^{2}$

AM = 5k

But, AM = MP + AP

Therefore, MP + AP = 5k --- (1)

It is given that MP - AP = 1 --- (2)

Multiply (1) and (2), we get.

(MP + AP)(MP - AP) = 5k × 1 = 5k

$MP^{2} -AP^{2} =5k$

Add and subtract $CP^{2}$ on the left side.

$MP^{2} + CP^{2} - CP^{2} -AP^{2} =5k$

$(MP^{2} + CP^{2}) - (CP^{2} +AP^{2}) =5k$ --- (3)

But, since CP ⊥ AM, Δ CMP and Δ CAP are right triangles. Therefore,

$MP^{2} +CP^{2} =CM^{2}$ and

$CP^{2} +AP^{2} =AC^{2}$

Now, (3) becomes,

$CM^{2} -AC^{2} =5k$

$(4k)^{2} -(3k)^{2} =5k$

$7k^{2} =5k$

7k = 5 or

$k=\frac{5}{7}$

AM = 5k

$=5(\frac{5}{7} )$

$=\frac{25}{7}$

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3 years ago

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Answer:

Step-by-step explanation:

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3 years ago

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kykrilka [37]

Explanation:

A logarithm is an exponent. An integer exponent tells you the number of times the base is a factor in the product:

100,000 = 10×10×10×10×10 = 10⁵

so, log₁₀(100,000) = log₁₀(10⁵) = 5, the exponent of the base.

You know that 10⁰ = 1 and 10¹ = 10. Since the logarithm is the exponent of 10, the log of a number between 1 and 10 is expected to be between 0 and 1.

You may also know that ...

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The number you're asking about, 5, has a logarithm that is about 0.69897.

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3 years ago