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3 years ago

I’m so confused on how to find the answer to this

Mathematics

1 answer:

nalin [4]3 years ago

4 0

Step-by-step explanation:

download an app called algebrator its way easier than asking brainly

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Can anybody help me plzz

Fittoniya [83]

Answer:

1 : 12 : 4

Step-by-step explanation:

Let A, G, C represent the ages of Alex, George, and Carl, respectively. We are told that ...

A = G +12

C = 3A . . . . . . . we read this as "3 times as old"; not "3 times older"

A + G + C = 68

Substituting for G and C, we have ...

A + (A -12) +(3A) = 68

5A = 80 . . . . . add 12

A = 16 . . . . . . . divide by 5

The other ages are then ...

G = A-12 = 16 -12 = 4

C = 3A = 3·16 = 48

The ratios are ...

G : C : A = 4 : 48 : 16

G : C : A = 1 : 12 : 4

7 0

3 years ago

(06.03 MC)

Burka [1]

Answer:

The answer is b, which is 3+x+3

8 0

2 years ago

Integration questions .

dlinn [17]

$\\\\\ \textbf{a)}\\\\~~~\displaystyle \int (6x- \sin 3x) ~ dx\\\\=6\displaystyle \int x ~ dx - \displaystyle \int \sin 3x ~ dx\\\\=6 \cdot \dfrac{x^2}2 - \dfrac 13 (- \cos 3x) +C~~~~~~~~~~~;\left[\displaystyle \int x^n~ dx = \dfrac{x^{n+1}}{n+1}+C,~~~n \neq -1\right]\\\\ =3x^2 +\dfrac{\cos 3x}3 +C~~~~~~~~~~~~~~~~~~~~;\left[\displaystyle \int \sin (mx) ~dx = -\dfrac 1m ~ (\cos mx)+C \right]\\$

$\textbf{b)}\\\\~~~~\displaystyle \int(3e^{-2x} +\cos (0.5 x)) dx\\\\=3\displaystyle \int e^{-2x} ~dx+ \displaystyle \int \cos(0.5 x) ~dx\\\\\\=-\dfrac 32 e^{-2x} + \dfrac 1{0.5} \sin (0.5 x) +C~~~~~~~~~~~~~~;\left[\displaystyle \int e^{mx}~dx = \dfrac 1m e^{mx} +C \right]\\\\\\=-\dfrac 32 e^{-2x} + 2 \sin(0.5 x) +C~~~~~~~~~~~~~~~~~;\left[\displaystyle \int \cos(mx)~ dx = \dfrac 1m \sin(mx) +C\right]\\\\\\=-1.5e^{-2x} +2\sin(0.5x) +C$

$\textbf{a)}\\\\y = \displaystyle \int \cos(x+5) ~ dx\\\\\text{Let,}\\\\~~~~~~~u = x+5\\\\\implies \dfrac{du}{dx} = 1+0~~~~~~;[\text{Differentiate both sides.}]\\\\\implies \dfrac{du}{dx} = 1\\\\\implies du = dx\\\\\text{Now,}\\\\y= \displaystyle \int \cos u ~ du\\\\~~~= \sin u +C\\\\~~~=\sin(x+5) + C$

$\textbf{b)}\\\\y = \displaystyle \int 2(5x-3)^4 dx\\\\\text{Let,}\\~~~~~~~~u = 5x-3\\\\\implies \dfrac{du}{dx} = 5~~~~~~~~~~;[\text{Differentiate both sides}]\\\\\implies dx = \dfrac{du}5\\\\\text{Now,}\\\\y = 2\cdot \dfrac 1 5 \displaystyle \int u^4 ~ du\\\\\\~~=\dfrac 25 \cdot \dfrac{u^{4+1}}{4+1} +C\\\\\\~~=\dfrac 25 \cdot \dfrac{u^5}5+C\\\\\\~~=\dfrac{2u^5}{25}+C\\\\\\~~=\dfrac{2(5x-3)^5}{25}+C$

$\textbf{a)}\\\\y = \displaystyle \int xe^{3x} dx\\\\\text{We know that,}\\\\ \displaystyle \int (uv) ~dx = u \displaystyle \int v ~ dx - \displaystyle \int \left[ \dfrac{du}{dx} \displaystyle \int ~ v ~ dx \right]~ dx\\\\\text{Let}, u =x~ \text{and}~ v=e^{3x} .\\\\y= \displaystyle \int xe^{3x} ~dx\\\\\\~~= x\displaystyle \int e^{3x} ~ dx - \displaystyle \int \left[\dfrac{d}{dx}(x) \displaystyle \int e^{3x}~ dx \right]~ dx\\\\\\$

$=x\displaystyle \int e^{3x}~ dx - \displaystyle \dfrac 13 \int \left(e^{3x} \right)~ dx\\\\\\=\dfrac{xe^{3x}}3 - \dfrac 13 \cdot \dfrac{ e^{3x}}3+C\\\\\\= \dfrac{xe^{3x}}{3}- \dfrac{e^{3x}}{9}+C\\\\\\=\dfrac{3xe^{3x}}{9}- \dfrac{e^{3x}}9 + C\\\\\\= \dfrac 19e^{3x}(3x-1)+C$

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8 0

2 years ago

The mean incubation time for a type of fertilized egg kept at 100.2100.2âÂ°f is 2323 days. suppose that the incubation times are

zlopas [31]

To solve this problem, what we have to do is to calculate for the z scores of each condition then find the probability using the standard normal probability tables for z.

The formula for z score is:

z = (x – u) / s

where,

x = sample value

u = sample mean = 23 days

s = standard deviation = 1 day

A. P when x < 21 days

z = (21 – 23) / 1

z = -2

Using the table,

P = 0.0228

Therefore there is a 2.28% probability that the hatching period is less than 21 days.

B. P when 23 ≥ x ≥ 22

P (z=-1) = 0.1587

z (x=23) = (23 – 23) / 1 = 0

P (z=0) = 0.5

P = 0.5 - 0.1587 = 0.3413

Therefore there is a 34.13% probability that the hatching period is between 22 and 23 days.

C. P when x > 25

z = (25 – 23) / 1

z = 2

P = 0.9772

This is not yet the answer since this probability refers to the left of z. Therefore the correct probability is:

P true = 1 – 0.9772

P true = 0.0228

<span>Therefore there is a 2.28% probability that the hatching period is more than 25 days.</span>

8 0

3 years ago

Let's call a song danceable if it has a danceability of at least 0.7. What is the probability that an individual song has a danc

vfiekz [6]

Answer:

32.54%

Step-by-step explanation:

To determine the probability that an individual song has a danceability of at least 0.7

Applying the excel formula below

=COUNTIF ( I2:I4771,">0.7") / ROWS( I2:I4771)

Hence ; The probability of an individual song having a danceability of at least 0.7 = 32,54%

3 0

3 years ago