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3 years ago

Which expression is equivalent to 24 + 64?

Mathematics

2 answers:

LUCKY_DIMON [66]3 years ago

8 0

It’s 3 brainliest me pls

Cerrena [4.2K]3 years ago

7 0

3. 8(3+8)

Explanation : 8x3 = 24 + 8x8 =64 which is congruent to 24 +64

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Find the circumcenter of triangle EFG with E(2,6) , F(2,4) , and G(6,4).

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The answer is (4,5) Mark me brainliest!!!!!

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3 years ago

HELP PLEASE >>>>>>>>>>>>>>>>>>>>>>

Damm [24]

Answer:

A. $\frac{2}{8}$

B. $\frac{6}{6}$

C. $\frac{8}{4}$

Step-by-step explanation:

For A pick a fraction that is less than a whole number.

For B pick a fraction that is a whole number.

For C pick a fraction that is greater than a whole number.

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3 years ago

A city planner wants to estimate, with a 99% confidence level, the average monthly residential water usage in the city. Based on

torisob [31]

Answer: 82

Step-by-step explanation:

The formula to find the sample size is given by :-

$n= (\dfrac{z^*\cdot \sigma}{E})^2$

, where E = Margin of error.

$\sigma$ = Population standard deviation for the prior study.

As per given , we have

$\sigma=387.40$ gallons.

Margin of error : E= 110 gallons

Critical value for 99% confidence level = z* = 2.576

Then, the required sample size : $n= (\dfrac{(2.576)\cdot (387.40)}{110})^2$

$n= (9.07220363636)^2\approx82$ {Rounded to the nearest integer.}

Hence, the minimum sample size = 82

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3 years ago

Factor x^2 -5x - 14 ??

zubka84 [21]

Answer:

=(x-7)(x+2)

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

The Pacific halibut fishery has been modeled by the differential equation dy dt = ky 1 − y K where y(t) is the biomass (the tota

Dafna1 [17]

Answer:

a. The biomass weighs 2.30 * 10^7 kg after a year

b. It'll take 2.56 years to get to 4*10^7kg

Step-by-step explanation:

k = 0.78,K = 6E7 kg

Given

dy/dt = ky(1- y/K)

Make ky dt the subject of formula

ky dt = dy/(1-y/K) --- make k dt the subject of formula

k dt = dy/(y(1-y/K))

k dt = K dy / y(K-y)

k dt = ((1/y) + (1/(K-y)))dy ---- integrate both sides

kt + c = ln(y/(K-y))

Ce^(kt) = y/(K-y)

Substitute the values of k and K

Ce^(0.78t) = y/(6*10^7 - y) ----- (1)

Given that y(0) = 2 * 10^7kg

(1) becomes

Ce^(0.78*0) = (2 * 10^7)/(6*10^7 - 2*10^7)

Ce° = (2*10^7)/(4*10^7

C = 2/7

Substitute 2/7 for C in (1)

2/7e^0.78t = y/(6*10^7 - y) ---(2)

We're to find the biomass a year later

So, t = 1

2/7e^0.78 = y/(6*10^7 - y)

0.62 = y/(6*10^7 - y)

y = 0.62(6*10^7 - y)

y = 0.62*6*10^7 - 0.62y

y + 0.62y = 0.62*6*10^7

1.62y = 0.62*6*10^7

1.62y = 3.72 * 10^7

y = 2.30 * 10^7kg.

Hence, the biomass weighs 2.30 * 10^7 kg after a year

Here, we're to calculate the time it'll take the biomass to get to 4*10^7 kg

Substitute 4*10^7 for y in (2)

2/7e^0.78t = 4*10^7/(6*10^7 - 4*10^7)

2/7e^0.78t = 4*10^7/2*10^7

2/7e^0.78t = 2

e^0.78t = (2*7)/2

e^0.78t = 2

t = 2 * 1/0.78

t = 2.56 years

Hence, it'll take 2.56 years to get to 4*10^7kg

8 0

3 years ago