Ask question

Ask question

All categories

3 years ago

15

How many terms are in the polynomial abcd + e - h27O twoO threeO fiveO six

1 answer:

Step2247 [10]3 years ago

8 0

3 terms. Abcd is one term
E is another term. -h27 is the third term

You might be interested in

to find the area of a trapezoid dylan uses the formula A=1/2 (B1+B2)h the bases have lengths of 3.6 cm and 12 1/3 cm. the height

Artemon [7]

Answer:

It is irrational because it can not be represented as a fraction of two integers

Step-by-step explanation:

Given

$B_1 = 3.6$

$B_2 = 12\frac{1}{3}$

$H = \sqrt[3]5$

Required

Why is the area irrational?

First, we need to calculate the area

$Area = \frac{1}{2}(B_1 + B_2) * H$

$Area = \frac{1}{2}(3.6 + 12\frac{1}{3}) * \sqrt[3]5$

$Area = \frac{1}{2}(3.6 + \frac{37}{3}) * 17100$

$Area = \frac{1}{2}(\frac{10.8+37}{3}) * 1.710$

$Area = \frac{1}{2}(\frac{47.8}{3}) * 1.7100$

$Area = \frac{47.8}{6} * 1.7100$

$Area = \frac{47.8* 1.7100}{6}$

$Area = \frac{81.738}{6}$

$Area = 13.623$

<em>It is irrational because it can not be represented as a fraction of two integers</em>

5 0

2 years ago

ANSWER ASAP RIGHT ANSWER = BRAINLIST 100 POINTS

ale4655 [162]

Answer:

-6.75

Hooe it helps!

Please mark brainliest

8 0

3 years ago

Read 2 more answers

PLS HELP WITH THIS MATH Equation

scZoUnD [109]

answer: (175+125)÷3

have a nice day

7 0

2 years ago

Use the exponential decay model, Upper A equals Upper A 0 e Superscript kt, to solve the following. The half-life of a certai

Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

$A=A_0 e^{kt}$

Where,

$A_0$ = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = $\frac{A_0}{2}$ ,

i.e.

$\frac{A_0}{2}=A_0 e^{19k}$

$\frac{1}{2} = e^{19k}$

$0.5 = e^{19k}$

Taking ln both sides,

$\ln(0.5) = \ln(e^{19k})$

$\ln(0.5) = 19k$

$\implies k = \frac{\ln(0.5)}{19}\approx -0.03648$

Now, if the substance to decay to 78% of its original amount,

Then $A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0$

$0.78 A_0=A_0 e^{-0.03648t}$

$0.78 = e^{-0.03648t}$

Again taking ln both sides,

$\ln(0.78) = -0.03648t$

$-0.24846=-0.03648t$

$\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7$

Hence, approximately the substance would be 78% of its initial value after 7 years.

5 0

3 years ago

Which is correct gelp

myrzilka [38]

Answer:

A

Step-by-step explanation:

8 0

2 years ago