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zhannawk [14.2K]
3 years ago
11

Help help! What are the answers!

Mathematics
1 answer:
vichka [17]3 years ago
8 0

Answer:

I have no idea what this is asking you to do bc it doesn't even make sense.

Step-by-step explanation:

(49-25)=8

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Please help me answer all the questions in this test!
Sholpan [36]

Answer:

4. 2(3k+4y-14).

5. 76+34.9.

6.10t-8

Step-by-step explanation:

4.since it's 6k+8y-14 all of the numbers can be divided by 2 then the you just put a 2 in front of everything and then(3k+4y-14).

5. (22+11.5x)+(54+23.4x).

=(22+54)+(11.5x+23.4x).

=76+34.9x.

=34.9x+76.

6.(9t-3)+(-5+t).

=(9t+t)+(-3-5).

=10t-8.

3 0
4 years ago
Help<br> Find the lateral area of this cone.<br> Round to the nearest tenth.
nata0808 [166]

Answer:

233.5 in²

Step-by-step explanation:

First we have to find the slant height.

5²+14²=c²

25+196=c²

221=c²

c=14.866

Now we can find the lateral area.

\pirl

\pi(5)(14.866)

=233.51566

To the nearest tenth=

233.5 in²

3 0
4 years ago
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What property do you use to solve (–5 • 7)25 = x(7 • 25)
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X=-5
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Write each measurement below in the table under an equivalent measure. Some
soldier1979 [14.2K]
1 yard=36 inches 1 foot=12 inches and 1/3 yard 1 inch= 1/36 yard and 1/12 foot
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Write an equation for an ellipse centered at the origin, which has foci at (\pm\sqrt{12},0)(± 12 ​ ,0)left parenthesis, plus min
lora16 [44]

Answer:

\mathbf{\dfrac{x^2}{49^2} +\dfrac{y^2}{37^2} =1}

Step-by-step explanation:

Given that :

the foci of the ellipse is (±√12,0) and C0-vertices are (0,±√37)

The foci are (-C,0) and (C ,0)

the focus has x-coordinates so the focus is  lie on x- axis.

The major axis also lie on x-axis

The minor axis lies on y-axis so C0-vertices are (0,±√37)

The given focus C = ae = √12

Given co-vertices ( minor axis) (0,±b) = (0,±√37)

b= √37

We can therefore express the  relation between the focus and semi major axes and semi minor axes as:

\mathbf{c^2 = a^2 - b^2 } \\ \\ \mathbf{a^2 = c^2 + b^2 } \\ \\ \mathbf{c^2 = ( \sqrt12)^2 - (\sqrt 37)^2 }  \\ \\ \mathbf{c^2 = 49 } \\ \\  \mathbf{c = \sqrt{49 }}

The equation of ellipse formula is:

\dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1

and we know that \mathbf{a=\sqrt{49}  \ \  and  \  \ b=\sqrt{37}}

Thus ; the equation of the ellipse at the origin is

\mathbf{\dfrac{x^2}{49^2} +\dfrac{y^2}{37^2} =1}

3 0
4 years ago
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