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scoundrel [369]
3 years ago
6

Which is a quadratic function? A. 3x + y^2 = 5 B. 3x^2 + y = 5 C. y = 3x + 5 D. x = 3y + 5

Mathematics
1 answer:
borishaifa [10]3 years ago
8 0

差は稀飲む、。サラダ等ノー。。。。。。お手)頬後へ襲おホミンゆー

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Y=3x+8 Y=-2x-2<br>can anyone help me with this ??
vivado [14]
Is it 2 different questions or just one?
6 0
3 years ago
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Work out 140-3^2×(8+4)+6÷2
kiruha [24]

Answer:

140 - 9 (12) + 3

140 - 108 +3

140 - 111

= 29

4 0
3 years ago
What’s the diameter of the circle of the circumference is 30 pi inches
ICE Princess25 [194]
30?
30pi = 2pi r
30pi / 2pi = 15
15 = r
double the radius to get the diameter
15)2 =30
6 0
4 years ago
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23. The letters of the alphabet are written on
Alla [95]

Answer:

21/676

Step-by-step explanation:

The total number of letters of the alphabet will be the total outcome which is 26 alphabets.

If consonant numbers are drawn out, the probability of drawing a consonant will be total consonants/total alphabets

Total consonants in the alphabet is 21.

Probability of drawing consonants = 21/26.

Since all the consonants are replaced before drawing a Z, our total outcome will not change i.e it will still be 26.

Probability of drawing a letter Z will be 1/26.

Therefore, the probability of drawing a consonant,replacing it, and then drawing a Z will be;

21/26×1/26

= 21/676

5 0
3 years ago
Let u = (1, 2 -2), v= (2, 1, 3) and w = j + 4k. a) Find cos theta, where theta is the angle in degrees between u and w.
vodka [1.7K]

Answer: The value of cos theta is -0.352.

Step-by-step explanation:

Since we have given that

\vec{u}=1\hat{i}+2\hat{j}-2\hat{k}\and\\\\\vec{w}=\hat{j}+4\hat{k}

We need to find the cos theta between u and w.

As we know the formula for angle between two vectors.

\cos\ \theta=\dfrac{\vec{u}.\vec{w}}{\mid u\mid \mid w\mid}

So, it becomes,

\cos \theta=\dfrac{2-8}{\sqrt{1^2+2^2+(-2)^2}\sqrt{1^2+4^2}}\\\\\cos \theta=\dfrac{-6}{\sqrt{17}\sqrt{17}}=\dfrac{-6}{17}=-0.352\\\\\theta=\cos^{-1}(\dfrac{-6}{17})=110.66^\circ

Hence, the value of cos theta is -0.352.

8 0
3 years ago
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