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frozen [14]
3 years ago
11

I really need help for this.

Mathematics
2 answers:
baherus [9]3 years ago
5 0
The graph increases by 0.1
ANEK [815]3 years ago
4 0
The graphs y-axis is increasing by 0.1 which makes small changes in rainfall seem more steep/dramatic than the real difference actually is.
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HELP!!!!!!!!!!!!!!!!!!!!!!!
Agata [3.3K]

Answer:

B would be (-1, -2)

C would be (-1, 1)

A would be (-4, -2)

Step-by-step explanation:

The ordered pairs are affected by the y value. The X value stays the same since you didn't move the shape any to the right or left. The y values went down three on all of the points because you moved the whole shape down three spots.

I hope this helps! Please mark as brainliest! Just ask if you need anymore help! :D Good Luck!

8 0
3 years ago
QUICKLY PLEASE
Greeley [361]

Answer:

the answer is d e and fans f

6 0
3 years ago
Read 2 more answers
The function h(x) is quadratic and h(3) = h(–10) = 0. Which could represent h(x)?
Lyrx [107]
I hope this helps you

6 0
3 years ago
Read 2 more answers
Given sintheta =7/11 and sectheta is less than 0, find costheta and tantheta
Korvikt [17]

Answer:

Step-by-step explanation:

Sin theta is the ratio of side opposite over hypotenuse of a reference angle situated at the origin in an x-y coordinate plane.  If sec theta is negative, then the only quadrant where sin is positive AND sec is negative is quadrant 2.  Remember that sec theta is the inverse of cos theta.  Puttling our right triangle in QII, the side measuring 7 is across from the angle and the hypotenuse is 11.  In order to find the cos theta and tan theta, we need the side adjacent to the angle.  Use Pythagorean's Theorem to find the side adjacent.

11^2=7^2+x^2 and

121=47+x^2 and

72=x^2 so

x=6\sqrt{2}

Remember that this value is why the sec is negative.  Because x is negative in QII, the cos theta is side adjacent over hypotenuse:

cos\theta=-\frac{6\sqrt{2} }{11}  and

tan\theta=-\frac{7}{6\sqrt{2} }

But we should probably rationalize that denominator, so

tan\theta=-\frac{7}{6\sqrt{2} }*\frac{\sqrt{2} }{\sqrt{2} } =-\frac{7\sqrt{2} }{12}

5 0
3 years ago
Plz help ill mark u brainliest
egoroff_w [7]

Answer:

<em>Part A:  3,</em>

<em>Part B:  - 4</em>

Step-by-step explanation:

<em>~ Part A ~ </em>

The problem would be in terms of such an expression: ( - 6 - ( - 3 )^2 )/ -5

Now knowing the expression, let us simplify in terms of basic algebra:

( -6 - 9 )/ -5 = ( - 15/ - 5 ) = <em>Answer; 3</em>

<em>~ Part B ~ </em>

We know the problem to be p^2/ r, but we must substitute the value of p being -10, and r ⇒ -25: ( - 10 )^2/ -25

Now let us simplify the expressions:

( - 10 )^2/ -25 = 100/ - 25 =<em> Answer; -4</em>

5 0
3 years ago
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