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3 years ago

7

I get none of this...please help?

1 answer:

mixer [17]3 years ago

7 0

1. The sum of the angles in a triangle is equal to 180. 3x+2x+x=180
2. 3x+2x+x=180 Combine like terms. 6x=180 180/6=30 x=30
3. 3*30=90; 2*30=60 1*30=30

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Please help me in so confused

jolli1 [7]

Answer:

A

Step-by-step explanation:

Similar figures have sides in the same ratio

169/1.3 = 17/x

x = 17×1.3/169

x = 0.131 m

5 0

3 years ago

Simplify the expression below and write it as a single logarithm:

OLEGan [10]

The simplification of 3log(x + 4) – 2log(x – 7) + 5log(x - 2) - log(x^2) is $\log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

<u>Solution:</u>

Given, expression is $3 \log (x+4)-2 \log (x-7)+5 \log (x-2)-\log \left(x^{2}\right)$

We have to write in as single logarithm by simplifying it.

Now, take the given expression.

$\rightarrow 3 \log (x+4)-2 \log (x-7)+5 \log (x-2)-\log \left(x^{2}\right)$

Rearranging the terms we get,

$\left.\rightarrow 3 \log (x+4)+5 \log (x-2)-2 \log (x-7)+\log \left(x^{2}\right)\right)$

$\text { since a } \times \log b=\log \left(b^{a}\right)$

$\rightarrow \log (x+4)^{3}+\log (x-2)^{5}-\left(\log (x-7)^{2}+\log \left(x^{2}\right)\right)$

$\text { We know that } \log a \times \log b=\log a b$

$\rightarrow \log \left((x+4)^{3} \times(x-2)^{5}\right)-\left(\log \left((x-7)^{2} \times\left(x^{2}\right)\right)\right.$

$\text { We know that } \log a-\log b=\log \frac{a}{b}$

$\rightarrow \log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

Hence, the simplified form $\rightarrow \log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

4 0

3 years ago

Read 2 more answers

When x+2 = 1/5y is written in standard form, what is the value or B

Ket [755]

Answer:

B=-1

Step-by-step explanation:

Standard form is Ax+By=C.

In this case, Ax is x, By is 1/5y, and C is +2.

To isolate the 2 (the C), subtract x from both sides of the equation:

x+2=1/5y

-x -x

2=-x+1/5y.

Now, the value of B, the x value, is -1.

4 0

3 years ago

The hydrogen ion concentration (h+) in a certain cleaning compound is (h+)=3.2x10^-11 use the formula ph=-log(h+) to find the pH

Furkat [3]

Answer:

The $pH$ of the clean compound is approximately 10.495.

Step-by-step explanation:

If the cleaning ion concentration of the cleaning compound is $[H^{+}] = 3.2\times 10^{-11}$ , then the $pH$ , no unit, of the compound is:

$pH = -\log_{10} [H^{+}]$ (1)

$pH = -\log_{10} (3.2\times 10^{-11})$

$pH = -\log_{10} 3.2 - \log_{10} (10^{-11})$

$pH = -\log_{10} \frac{32}{10} +11$

$pH = -\log_{10} 32 + \log_{10} 10 + 11$

$pH = -\log_{10} 32 +12$

$pH \approx 10.495$

The $pH$ of the clean compound is approximately 10.495.

6 0

2 years ago

Whoever guesses not only the song but the singer will get brainliest if there are any absurd answers they will be deleted.

ivanzaharov [21]

Answer:

Somebodys problem I think by Morgan something I don’t know

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers