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natka813 [3]
3 years ago
14

Mason claims that y= 3 x2 -6 is a function, but not a linear one. Select the statement that supports Mason's claim.

Mathematics
2 answers:
11111nata11111 [884]3 years ago
6 0
D. Cause it does not contain the point
denpristay [2]3 years ago
3 0

Answer:

D. The function does not contain the point (0,0).

Step-by-step explanation:

You might be interested in
The recipe calls for 4 cups of strawberries for every 2 cups of whipped cream. Write this comparison as a ratio.
Gwar [14]

A ratio is a comparison of two things. It can be written with a colon, like 5:3, as a fraction, like 5/3, or with words, like 5 to 3. Our ratio is strawberries:whipped cream. If we substitute in our numerical values, we get:

4:2

However, this ratio is not simplified. To simplify it, we must divide both the left and right sides of the ratio by their shared greatest common factor, or GCF. To find their GCF, let's look at the factors of both numbers. 4 - 1, 2, 4. 2 - 1,2.

The greatest shared factor is 2, so we need to divide both parts of the ratio by 2, as follows:

4:2 = 4/2 : 2/2 = 2:1

Therefore, your answer is 4:2 or when simplified, 2:1.

Hope this helps!

3 0
4 years ago
I need help pweaseeee :pppp
Nookie1986 [14]

Answer:

TELL ME WHY I AM FOCUSED ON THE OTHER PICS MORE THAN THE PROBLEMS

7 0
3 years ago
Find the values of c such that the area of the region bounded by the parabolas
evablogger [386]
For there to be a region bounded by the two parabolas, you first need to find some conditions on c. The two parabolas must intersect each other twice, so you need two solutions to

16x^2-c^2=c^2-16x^2

You have

32x^2=2c^2\implies 16x^2=c^2\implies x=\pm\dfrac{|c|}4

which means you only need to require that c\neq0. With that, the area of any such bounded region would be given by the integral

\displaystyle\int_{-4|c|}^{4|c|}\bigg((c^2-16x^2)-(16x^2-c^2)\bigg)\,\mathrm dx

since c^2-16x^2>16x^2-c^2 for all c\neq0. Now,

\displaystyle\int_{-|c|/4}^{|c|/4}\bigg((c^2-16x^2)-(16x^2-c^2)\bigg)\,\mathrm dx=2\int_0^{|c|/4}(2c^2-32x^2)\,\mathrm dx

by symmetry across the y-axis. Integrating yields

\displaystyle2\int_0^{|c|/4}(2c^2-32x^2)\,\mathrm dx=4\int_0^{|c|/4}(c^2-16x^2)\,\mathrm dx
=4\left[c^2x-\dfrac{16}3x^3\right]_{x=0}^{x=|c|/4}
=c^2|c|-\dfrac{|c|^3}3
=\dfrac{2|c|c^2}3=144
|c|c^2=216

Since 216=6^3, you have c=\pm6.
3 0
3 years ago
Please help me with this question.<br> Will mark brainliest <br> Thanks so much
Alina [70]

Answer:

☑ 30y²

☑ 30y² + x

Step-by-step explanation:

Polynomials contain indeterminates [variables] and operation performances, non-including negative exponents, fractional exponents, etcetera.

I am joyous to assist you anytime.

6 0
4 years ago
Subtract 386 from 72,000
goldfiish [28.3K]

71614 is the answer, this can be found by simply solving this expression: 72000 - 386

6 0
3 years ago
Read 2 more answers
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