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3 years ago

Solve for B. A=3B+7C

Mathematics

1 answer:

Archy [21]3 years ago

6 0

Hey there!!

Given:

... A=3B+7C

Subtracting 7C on both sides:

... A-7C=3B

Dividing by B on both sides:

... (A-7C)/3=B

<em>Hope my answer helps!</em>

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Select all the expressions that are equivalent to −5/6 / -1/3

Sindrei [870]

Step-by-step explanation:

We need to find an expression for $\dfrac{\dfrac{-5}{6}}{\dfrac{-1}{3}}$ .

We can solve it as follows.

We know that,

$\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}$

So,

$\dfrac{\dfrac{-5}{6}}{\dfrac{-1}{3}}=\dfrac{-5}{6}\times -3\\\\=\dfrac{5}{2}$

$=2\dfrac{1}{2}$

Hence, this is the required solution.

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2 years ago

3 6/4 in simplest form

ivanzaharov [21]

Answer:

4 1/2.

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6 0

2 years ago

Which logarithmic equation is equivalent to the exponential equation below?

aleksandr82 [10.1K]

Option B: $\ln 6=5 x$ is the correct answer.

Explanation:

The exponential equation is $e^{5 x}=6$

If $f(x)=g(x)$ , then $\ln (f(x))=\ln (g(x))$

Thus, the equation becomes

$\ln \left(e^{5 x}\right)=\ln (6)$

Applying log rule, $\log _{a}\left(x^{b}\right)=b \cdot \log _{a}(x)$ and thus the equation becomes

$5 x \ln (e)=\ln (6)$

Since, we know that, $\ln (e)=1$ , using this we get,

$5 x=\ln (6)$

Hence, the logarithmic equation which is equivalent to the exponential equation $e^{5 x}=6$ is $\ln 6=5 x$

Thus, Option B is the correct answer.

7 0

3 years ago

2/5 ÷ −13/11 ≈ pls help me pls

LenaWriter [7]

Answer: -22/65

Step-by-step explanation: PLEASE GIVE BRAINLIEST IT HElPS ALOT

8 0

2 years ago

Read 2 more answers

Consider the set of differences, denoted with d, between two dependent sets: 84, 85, 83, 63, 61, 100, 98. Find the sample standa

sveta [45]

Answer:

The sample standard deviation is 15.3.

Step-by-step explanation:

Given data items,

84, 85, 83, 63, 61, 100, 98,

Number of data items, N = 7,

Let x represents the data item,

Mean of the data points,

$\bar{x}=\frac{84+85+83+63+61+100+98}{7}$

$=82$

Hence, sample standard deviation would be,

$\sigma= \sqrt{\frac{1}{N-1}\sum_{i=1}^{N} (x_i-\bar{x})^2}$

$=\sqrt{\frac{1}{6}\sum_{i=1}^{7} (x_i-82)^2}$

$=\sqrt{\frac{1}{6}\times 1396}$

$=\sqrt{232.666666667}$

$=15.2534149182$

$\approx 15.3$

4 0

2 years ago