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2 years ago

When the rectangle ABCD is reflected across line EF, the image is DCBA. How do you

Mathematics

2 answers:

Simora [160]2 years ago

3 0

The answer is D I took it

Elina [12.6K]2 years ago

3 0

D!! IS THE ANSWERRRRR:)

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Evaluate.

Daniel [21]

Answer:

d) 192

Step-by-step explanation:

Your calculator can do this for you.

Or, you can simplify to

$12\cdot\dfrac{4^{-2}}{4^{-4}}=12(4^{-2+4})=12(4^2)=192$

3 0

3 years ago

Find the Measure of the missing angles.

Savatey [412]

Step-by-step explanation:

1= 118°+angle 1=180°

angle 1= 180°-118°=62°

angle 2=180°-135°=45°

angle 3=135°

8 0

3 years ago

Solve for yyy. -3 = 7\left(y-\dfrac97\right)−3=7(y− 7 9 )minus, 3, equals, 7, left parenthesis, y, minus, start fraction, 9, d

Leni [432]

Answer:

y = 6/7

Step-by-step explanation:

The equation given is:

-3 = 7(y - 9/7)

To solve this, we will expand the bracket and afterwards, collect like terms:

-3 = (7*y) - (9/7 * 7)

-3 = 7y - 9

Collecting numeric terms to the left hand side:

-3 + 9 = 7y

=> 6 = 7y

=> y = 6/7

That is the solution.

3 0

3 years ago

Read 2 more answers

Round 36,468,212 to the nearest thousand.

just olya [345]

Answer: 36,468,000

Step-by-step explanation: can I have brainliest please

8 0

2 years ago

The state highway department is studying traffic patterns on one of the busiest highways in the state. As part of the study, the

larisa86 [58]

Answer:

The need to sample 76 days.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.005 = 0.995$ , so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How many days do they need to sample

We need to sample n days.

n is found when M = 300.

We have that $\sigma = 1010$

$M = z*\frac{\sigma}{\sqrt{n}}$

$300 = 2.575*\frac{1010}{\sqrt{n}}$

$300\sqrt{n} = 2.575*1010$

$\sqrt{n} = \frac{2.575*1010}{300}$

$(\sqrt{n})^{2} = (\frac{2.575*1010}{300})^{2}$

$n = 75.1$

Rounding up

The need to sample 76 days.

3 0

2 years ago