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kiruha [24]
3 years ago
12

Jada’s teacher ask her to write a three-digit number that uses the digit 4 once and the digit 7 twice. The value of one 7 needs

to be 1/10 the value of the other 7. What number can jada write?
Mathematics
1 answer:
viva [34]3 years ago
6 0
477

The value of the first 7 from the left is 70.

If you divide 70 by 10, you get the value of the second seven, 7.

Another way to know this is equivalent fractions:

1/10 = 7/10

10 X 7 = 70
1 X 7 = 7
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2 years ago
Let $f(x) = 2x^2 + 3x - 9,$ $g(x) = 5x + 11,$ and $h(x) = -3x^2 + 1.$ Find $f(x) - g(x) + h(x).$
Viefleur [7K]

QUESTION 1

Given that:

f(x)=2x^2+3x-9,

g(x)=5x+11,

and

h(x)=-3x^2+1

Then;

f(x)-g(x)+h(x)=2x^2+3x-9-(5x+11)+(-3x^2+1)

f(x)-g(x)+h(x)=2x^2+3x-9-5x-11-3x^2+1

Group similar terms;

f(x)-g(x)+h(x)=2x^2-3x^2+3x-5x-11-9+1

Simplify;

f(x)-g(x)+h(x)=-x^2-2x-19

QUESTION 2

Given that;

f(x)=4x-7.

g(x)=(x+1)^2

and

s(x)=f(x)+g(x)

Substitute the functions;

s(x)=4x-7+(x+1)^2

Substitute x=3

s(3)=4(3)-7+(3+1)^2

s(3)=12-7+(4)^2

s(3)=5+16

s(3)=21

QUESTION 3

Given:

f(x)=3x+2

g(x)=x^2-5x-1

f(g(x))=f(x^2-5x-1)

This implies that;

f(g(x))=3(x^2-5x-1)+2

Expand the parenthesis;

f(g(x))=3x^2-15x-3+2

f(g(x))=3x^2-15x-1

QUESTION 4

The given function is;

f(x)=3(x-6)^2+1

Let

y=3(x-6)^2+1

\Rightarrow y-1=3(x-6)^2

\Rightarrow \frac{y-1}{3}=(x-6)^2

\Rightarrow \sqrt{\frac{y-1}{3}}=x-6

\Rightarrow x=6+\sqrt{\frac{y-1}{3}}

The range is:

\frac{y-1}{3}\ge0

y-1\ge0

y\ge1

The interval notation is;

[1,+\infty)

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Answer:

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Step-by-step explanation:

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