All categories

2 years ago

Can someone actually help and not delete their answer I rlly need your help I’ll make u brainliest or whatever

Mathematics

1 answer:

Brrunno [24]2 years ago

5 0

Answer:

p = -30 + 20

Hope this helps!

You might be interested in

QUICK PLEASE. Calculate the final amount if $10 000 invested for 6 years at 4% compounded quarterly.

Nikolay [14]

The final amount if $10 000 invested for 6 years at 4% compounded quarterly is $12,697

<h3>Compound interest</h3>

The formula for calculating the compound interest is given as:

A = P(1+r/n)^nt

Given the following parameters

A = 10000(1+0.04/4)^4(6)

A = 10000(1+0.01)^24

A = 10000(1.01)^24

A = 1.2697(10,000)

A = $12,697

Hence the final amount if $10 000 invested for 6 years at 4% compounded quarterly is $12,697

Learn more on compound interest here: brainly.com/question/24924853

#SPJ1

7 0

2 years ago

Which of the following are valid probability distributions?

Anvisha [2.4K]

Answer:

Step-by-step explanation:

6 0

3 years ago

PLEASE HELP ITS URGENT. I’m not sure how to do this!! I will mark brainlist!!

elena-s [515]

Answer:

2y+3=5y-6. (Since all sides of rhoumbus are equal)

2y-5y=-6-3

-3y=-9

y=-9/-3

y=3

I hope it helps!!

8 0

3 years ago

How to solve this question

yarga [219]

Answer:

BC = 3.6

Step-by-step explanation:

Since the triangles are similar then the ratios of corresponding sides are equal, that is

$\frac{BC}{DE}$ = $\frac{AC}{AE}$ , substitute values

$\frac{BC}{1.2}$ = $\frac{4.5}{1.5}$ ( cross- multiply )

1.5 BC = 5.4 ( divide both sides by 1.5 )

BC = 3.6

6 0

3 years ago

Root 3 cosec140° - sec140°=4<br>prove that<br><br>

Lorico [155]

Answer:

Step-by-step explanation:

We are to show that $\sqrt{3} cosec140^{0} - sec140^{0} = 4\\$

<u>Proof:</u>

From trigonometry identity;

$cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}$

$\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\$

From trigonometry, 2sinAcosA = Sin2A

$= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\= \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2 \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}$

Also note that sin(B-C) = sinBcosC - cosBsinC

sin420cos140 - cos420sin140 = sin(420-140)

The resulting equation becomes;

$\frac{4(sin(420-140)) }{sin280}$

= $\frac{4sin280}{sin280}\\ = 4 \ Proved!$

3 0

3 years ago