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never [62]
2 years ago
15

Marina correctly simplified the expression StartFraction negative 4 a Superscript negative 2 Baseline b Superscript 4 Baseline O

ver 8 a Superscript negative 6 Baseline b Superscript negative 3 Baseline EndFraction, assuming that a not-equals 0, b not-equals 0. Her simplified expression is below.
Negative one-half a Superscript 4 Baseline b Superscript empty box

What exponent should Marina use for b?
Mathematics
1 answer:
Temka [501]2 years ago
5 0

The exponent Marina should use for b is: D. 7.

<h3>How to Divide Exponents with the same Base?</h3>

When dividing, you subtract the exponents. I.e. a^m \div a^n = a^{m - n}.

Using the rule, we would also subtract the exponents of b in the simplified expression. Thus:

b^4 \div b^{-3}

Subtract the exponents

4 -(-3) = 4 + 3 = 7.

Therefore, the Maria should use the exponent b = 7.

Learn more about dividing exponents on:

brainly.com/question/2263967

#SPJ1

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What is 4(x−2)=3(2y−1) in standard form?
exis [7]

Answer: 4x-6y=5

Step-by-step explanation: Using the formula, Ax+By=C, write the equation in standard form.

Hope this helps you out! ☺

8 0
3 years ago
Can you help me write this in standard form <br><img src="https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B3%7D%7B2%7D%20x%20-%20%
Karo-lina-s [1.5K]
I think it’s 15x-10y=2
3 0
3 years ago
Nachelle is 1.45 meters tall. At 11 a.m., she measures the length of a tree's shadow to be 23.05 meters. She stands 18.9 meters
GREYUIT [131]

Step-by-step explanation:

this creates 2 similar triangles.

that means all angles are the same. and all the side lengths of one triangle correlate to the corresponding side lengths of the other triangle by the same multiplication factor.

her shadow is 23.05 - 18.9 = 4.15 m long.

the factor between the shadow lengths is then

4.15 × f = 23.05

f = 23.05 / 4.15 = 5.554216867...

the same factor now applies to the relation of the heights :

1.45 × 5.554216867... = 8.053614458... m

rounded the tree is 8.05 m tall.

5 0
2 years ago
A gallup survey indicated that 72% of 18- to 29-year-olds, if given choice, would prefer to start their own business rather than
Neko [114]

Answer:

The probability that no more than 70% would prefer to start their own business is 0.1423.

Step-by-step explanation:

We are given that a Gallup survey indicated that 72% of 18- to 29-year-olds, if given choice, would prefer to start their own business rather than work for someone else.

Let \hat p = <u><em>sample proportion of people who prefer to start their own business</em></u>

The z-score probability distribution for the sample proportion is given by;

                               Z  =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, p = population proportion who would prefer to start their own business = 72%

            n = sample of 18-29 year-olds = 600

Now, the probability that no more than 70% would prefer to start their own business is given by = P( \hat p \leq 70%)

       P( \hat p \leq 70%) = P( \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } \leq \frac{0.70-0.72}{\sqrt{\frac{0.70(1-0.70)}{600} } } ) = P(Z \leq -1.07) = 1 - P(Z < 1.07)

                                                                       = 1 - 0.8577 = <u>0.1423</u>

The above probability is calculated by looking at the value of x = 1.07 in the z table which has an area of 0.8577.

3 0
3 years ago
If
Leno4ka [110]

Answer:

\frac{s^2-25}{(s^2+25)^2}

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given:\mathcal{L}[t \cos 5t]=(-1)F'(s) with F(s)=\mathcal{L}[\cos 5t].

Now, F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt. Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt.

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s).

Solving for F(s) on the last equation, F(s)=\frac{s}{s^2+25}, then the Laplace transform we were searching is -F'(s)=\frac{s^2-25}{(s^2+25)^2}

3 0
3 years ago
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