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3 years ago

Ill give brainly :))))

Mathematics

2 answers:

trasher [3.6K]3 years ago

7 0

To simplify the expression all you must do is add common numbers, when you add all the negatives you are left with -24 and 4r.
So I believe the answer is
4r + -24
or
4r-24

Luden [163]3 years ago

3 0

20 now give brainly

if that is not it its -20

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Anyone got answers need help

Mkey [24]

∠DCE = ∠BAE because of alternate interior ∠s

∠CDE = ∠ABE because of alternate interior ∠s

Triangle DCE is congruent to Triangle BAE because of ASA

AE = CE because of CPCTC

BE = DE because of CPCTC

4 0

3 years ago

Prove that Tan x + Cos x = 1 /Sin x .Cos x

Anit [1.1K]

Answer:

tanx + cotx = $\frac{1}{sinxcosx}$

Step-by-step explanation:

Explanation

Take R.H.S = $\frac{1}{sinxcosx}$

= $\frac{sin^{2}x+cos^{2}x }{sinxcosx}$

= $\frac{sin^{2}x }{sinx cosx} + \frac{cos^{2} x}{sinxcosx}$

= $\frac{sinx}{cosx} + \frac{cosx}{sinx}$

= tanx + cotx

tanx + cotx = $\frac{1}{sinxcosx}$

5 0

3 years ago

PLASS HELPPP WILL GIVE BRAINLISTTT DUE RN

scoray [572]

Answer:

The area of the rectangle is 0.0875 sq ft.

Step-by-step explanation:

Given that,

The length of the rectangle, L = 0.35 feet

The breadth of the rectangle, B = 1/4 feet

We need to find the area of the rectangle. The formula for the area of a rectangle is given by :

A = LB

Put all the values,

$A=\dfrac{1}{4}\times 0.35\\\\A=0.0875\ ft^2$

So, the area of the rectangle is 0.0875 sq feet.

4 0

2 years ago

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc

Bingel [31]

Answer:

We need a sample size of at least 719

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:

$\alpha = \frac{1-0.95}{2} = 0.025$

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 $1-0.025 = 0.975$ , so $z = 1.96$

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 large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when $M = 0.3, \sigma = 4.103$ . So

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

$0.3 = 1.96*\frac{4.103}{\sqrt{n}}$

$0.3\sqrt{n} = 1.96*4.103$

$\sqrt{n} = \frac{1.96*4.103}{0.3}$

$(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}$

$n = 718.57$

Rouding up

We need a sample size of at least 719

6 0

3 years ago

Val walks 2 3 fifths miles each day. Bill runs 10 miles once every 4 days. In four days, who covers the greater distance?

melisa1 [442]

Val covers greater distance because 2 3/5 x 4 = 10 2/5. Bill only ran 10 miles so Val covered 2/5 more miles than him.

3 0

3 years ago