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Dominik [7]
3 years ago
15

Whic graph shows y=x^2-3 ?

Mathematics
1 answer:
Snowcat [4.5K]3 years ago
4 0
The answer is A because pintasan y is -3
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The two consecutive odd integers are 57 & 59.
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3 years ago
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I need help asap ! this is 6th grade math surface area .
faltersainse [42]

Answer:

<h2>S.A. = 62</h2>

Step-by-step explanation:

We have:

two rectangles 2 × 3

two rectangles 3 × 5

two rectangles 2 × 5

Calculate the areas:

A₁ = (2)(3) = 6

A₂ = (3)(5) = 15

A₃ = (2)(5) = 10

The Surface Area:

S.A. = 2A₁ + 2A₂ + 2A₃

S.A. = (2)(6) + (2)(15) + (2)(10) = 12 + 30 + 20 = 62

8 0
3 years ago
MZJ and m M are base angles of isosceles trapezoid JKLM. If mZJ = 18x + 8 and m_M=11x + 15 find m2K.
maxonik [38]

The question is incomplete. Here is the complete question.

m∠J and m∠Kare base angles of an isosceles trapezoid JKLM.

If m∠J = 18x + 8, and m∠M = 11x + 15 , find m∠K.

A. 1

B. 154

C. 77

D. 26

Answer: B. m∠K = 154

Step-by-step explanation: <u>Isosceles</u> <u>trapezoid</u> is a parallelogram with two parallel sides, called Base, and two non-parallel sides that have the same measure.

Related to internal angles, angles of the base are equal and opposite angles are supplementary.

In trapezoid JKLM, m∠J and m∠M are base angles, so they are equal:

18x + 8 = 11x + 15

7x = 7

x = 1

Now, m∠K is opposite so, they are supplementary, which means their sum results in 180°:

m∠J = 18(1) + 8

m∠J = 26

m∠K + m∠J = 180

m∠K + 26 = 180

m∠K = 154

The angle m∠K is 154°

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3 years ago
ASAP I REALLY NEED HELP I WILL MAKE BRAINLIEST
svetlana [45]

Answer:

90

Step-by-step explanation:

6 0
3 years ago
Write an equation in standard form of the hyperbola described.
marishachu [46]

Check the picture below, so the hyperbola looks more or less like so, so let's find the length of the conjugate axis, or namely let's find the "b" component.

\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill

\begin{cases} h=0\\ k=0\\ a=2\\ c=4 \end{cases}\implies \cfrac{(x-0)^2}{2^2}-\cfrac{(y-0)^2}{b^2} \\\\\\ c^2=a^2+b^2\implies 4^2=2^2+b^2\implies 16=4+b^2\implies \underline{12=b^2} \\\\\\ \cfrac{(x-0)^2}{2^2}-\cfrac{(y-0)^2}{12}\implies \boxed{\cfrac{x^2}{4}-\cfrac{y^2}{12}}

5 0
2 years ago
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