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lidiya [134]
3 years ago
12

I need help on this question. Write an equivalent expression. (x + 7) + 3y

Mathematics
2 answers:
Mandarinka [93]3 years ago
6 0
(x+7)+3y = 
x+7+3y = 
2x+14+6y
s2008m [1.1K]3 years ago
6 0
(7+7) + 3(2) x=7 y= 2 The equation equals 20
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Is 100 a good estimate for the difference of 712 and 589? If it is, explain why it is a good estimate. If it is not, explain why
Masja [62]
It is a bad estimate because the gap between 712 and 589 is clearly (if you were just estimating in hundreds) would be 200

200 would be a better guess
4 0
4 years ago
What is 8x8x3to the second power
sleet_krkn [62]

Answer: If you're talking about the answer to 8*8*3 (192) to the second power, the answer is 36,864. If you're talking about 8*8*3^2 (with the three being the number with the raised power), the answer is 576

Step-by-step explanation:

8x8x3=192

192^2=192x192= 36,864

8x8x3^2=576

6 0
3 years ago
Read 2 more answers
In a G.P., the third term exceeds the first term by 16.
Tresset [83]

Answer:

Step-by-step explanation:

In a G.P, the nth term is given as

Un=ar^(n-1)

Where

a is first term

n is nth term

And r is common ratio

So in the question given above,

The third term exceed the first term by 16

i.e U3=U1+16

Where U1=a. First term

U3=a+16

Given also that, the sum if the third term and fourth term is 72.

Then U3+U4=72.

We are told to find common ratio (r)

U3=ar^3-1

U3=ar^2

Also, U4=ar^3

U3+U4=72

ar^2+ar^3=72

ar^2(1+r)=72. equation 1

Also for

U3=a+16

ar^2=a+16

ar^2-a=16

a(r^2-1)=16. From (x^2-y^2)=(x+y)(x-y)

Then,

a(r-1)(r+1)=16. Equation 2

Divide equation 2 by equation 1

a(r-1)(r+1)/ar^2(1+r) =16/72

Then a cancel a and (1+r) cancel (1+r)

So,

(r-1)/r^2=2/9

Cross multiply

9(r-1)=2r^2

2r^2-9r+9=0

Solving the quadratic equation

2r^2-6r-3r+9=0

2r(r-3)-3(r-3)=0

(r-3)(2r-3)=0

r-3=0. Or. 2r-3= 0

Then r=3 or r=3/2

8 0
3 years ago
What is the range of the function f(x)=1/2sqrt of x ?
Makovka662 [10]

Answer:

All real numbers greater than or equal to 0.

Step-by-step explanation:

x must be greater or equal to zero because there is no real square root of a negative number.

So the lowest value of the function is 0 and  it can be any value greater than that.

8 0
4 years ago
A population is normally distributed with mean 18 and standard deviation 1.7. (a) Find the intervals representing one, two, and
elena55 [62]

Answer:

Interval within 1 standard deviation: (16.3,19.7)

Interval within 2 standard deviation: (14.6, 21.4)

Interval within 3 standard deviation: (12.9, 23.1)

Step-by-step explanation:

We are given the following in the question:

Population mean, \mu = 18

Standard deviation, \sigma = 1.7

We have to find the following intervals:

Interval within 1 standard deviation:

\mu \pm 1\sigma\\=18 \pm 1.7\\=(16.3, 19.7)

Interval within 2 standard deviation:

\mu \pm 2\sigma\\=18 \pm 2(1.7)\\= 18 \pm 3.4\\=(14.6,21.4)

Interval within 3 standard deviation:

\mu \pm 3\sigma\\=18 \pm 2(1.7)\\= 18 \pm 5.1\\=(12.9,23.1)

6 0
4 years ago
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