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makvit [3.9K]
3 years ago
12

Determine whether the lines l1 and l2 given by the vector equations are parallel, perpendicular, or neither. L1: r(t) = (-2 + 4t

)i + (2 + t)j l2: r(s) = (3 + 3s)i + (2 - 12s)j
Mathematics
1 answer:
ss7ja [257]3 years ago
4 0

Over which interval(s) is the function decreasing?

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Marissa has a photograph that measures 2 in. by 4 in. She has mounted the picture on a mat so that there is a border that measur
DanielleElmas [232]

Step-by-step explanation:

The initial image of the photo is 2 in by 4 in.  The mat is 4 in by 6 in.

The new image is dilated by a scale of 2.  So we double the dimensions.  The new photo is 4 in by 8 in.  The new mat is 8 in by 12 in.

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What is the value of x?<br> O 2<br> 03<br> 06<br> 07<br> "Е
Svet_ta [14]
Ima pretty sure it’s 06
6 0
2 years ago
4. Solve x -- 7 = 25.<br> A 18<br> B 32
gregori [183]

Answer:

B

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
| 2x + 2y = 10<br> y = 5+x
melisa1 [442]
2x + 2(5+x)=10
2x+10+2x=10
4x+10=10
4x=0
X=0
6 0
3 years ago
what is the polynomial function of lowest degree with leading coefficient of 1 and roots 2 and 1 + square root2
Marta_Voda [28]
If the roots to such a polynomial are 2 and 1+\sqrt2, then we can write it as

(x-2)(x-(1+\sqrt2))

courtesy of the fundamental theorem of algebra. Now expanding yields

(x-2)(x-1-\sqrt2)=x^2-2x-(1+\sqrt2)x+2(1+\sqrt2)=x^2-(3+\sqrt2)x+2+2\sqrt2

which would be the correct answer, but clearly this option is not listed. Which is silly, because none of the offered solutions are *the* polynomial of lowest degree and leading coefficient 1.

So this makes me think you're expected to increase the multiplicity of one of the given roots, or you're expected to pull another root out of thin air. Judging by the choices, I think it's the latter, and that you're somehow supposed to know to use 1-\sqrt2 as a root. In this case, that would make our polynomial

(x-2)(x-(1+\sqrt2))(x-(1-\sqrt2))=x^3-4x^2+3x+2

so that the answer is (probably) the third choice.

Whoever originally wrote this question should reevaluate their word choice...
5 0
3 years ago
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