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Allushta [10]
3 years ago
7

Can someone help me with this

Mathematics
1 answer:
Morgarella [4.7K]3 years ago
6 0

Answer:

FG=30

Step-by-step explanation:

Since we know that Point G is on the Segment FH, it doesn't really matter where G is, but we can know for certain that:

FH=FG+GH

We are given that FH is 4x, GH is x, and FG is 2x+10. Substitute:

4x=(2x+10)+x

Solve for x. On the right, combine like terms:

4x=3x+10

Subtract 3x from both sides:

x=10

So, the value of x is 10.

To find the value of FG, substitute 10 into its x:

FG=2x+10\\FG=2(10)+10

Multiply:

FG=20+10

Add:

FG=30

And we're done!

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Which of the following could be f(x)?
NikAS [45]

Answer:

f(x) = -2x² + 2

Step-by-step explanation:

If you remember the y = x² parent graph, you will be able to derive your equation from visualization.

The vertex of this graph would be at (0, 2), so your degree coefficient would be negative. Your y-intercept would also be at (0, 2). Since both 1 and -1 are included, you can tell it is a parabola, and hence the 3rd option is your answer.

8 0
3 years ago
The Dulac Box plant produces 500 cypress packing boxes in two-10 hour shifts. Due to higher demand, Dulac has decided to operate
Keith_Richards [23]

Answer:

Production has increased 20/day

Step-by-step explanation:

In the first scenario production is 500/day and productivity 25 set/hour. After the changes, production is 600/day and 25 set/hour.

So productivity remains the same, nevertheless, as there are more productive hours per day, production raises, in this case the can be calculated as (New Production-Old Production)/Old Production=(600-500)/500=100/500=0.2=20%.

As productivity remains the same, you do not ge more sets/shift, as shifts are shorter (8 instead of hours, so you get 200/shift instead of 250/shift). The rest of the option is false as productivity remains constant

6 0
4 years ago
3/5 to second power times 25
Anestetic [448]
The answer is 9 because 3/5 to second power is 9/25 then you multiply it by 25 and you get 9
7 0
3 years ago
Describe the end behavior of a 14th degree polynomial with a positive leading coefficient.
DiKsa [7]

<u><em>Answer:</em></u>

1)

f(x)→ ∞ when x→∞ or x→ -∞.

2)

when  x→ ∞ then f(x)→ -∞

        and when x→ -∞ then f(x)→ ∞

<u><em>Step-by-step explanation:</em></u>

<em>" The </em><em>end behavior</em><em> of a polynomial function is the behavior of the graph of as approaches positive infinity or negative infinity. The degree and the leading coefficient of a polynomial function determine the end behavior of the graph "</em>

1)

a 14th degree polynomial with a positive leading coefficient.

Let f(x) be the polynomial function.

Since the degree is an even number and also the leading coefficient is positive so when we put negative or positive infinity to the function i.e. we put x→∞ or x→ -∞ ; it will always lead the function to positive infinity

i.e. f(x)→ ∞ when x→∞ or x→ -∞.

2)

a 9th degree polynomial with a negative leading coefficient.

As the degree of the polynomial is odd and also the leading coefficient is negative.

Hence when x→ ∞ then f(x)→ -∞ since the odd power of x will take it to positive infinity but the negative sign of the leading coefficient will take it to negative infinity.

When x→ -∞ then f(x)→ ∞; since the odd power of x will take it to negative infinity but the negative sign of the leading coefficient will take it to positive infinity.

Hence, when  x→ ∞ then f(x)→ -∞

        and when x→ -∞ then f(x)→ ∞



8 0
3 years ago
Y+4−1=18 i need help plz
d1i1m1o1n [39]

Answer:

y=15

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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