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2 years ago

Subtract. (8u - 2v + 7w) and (5u + v + 6w)

Mathematics

1 answer:

KiRa [710]2 years ago

5 0

Answer:

c)3u-3v-w

Step-by-step explanation:

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Triangle RST was dilated by a scale factor of One-half. The image, triangle R'S'T', is an isosceles triangle, with each leg meas

andrew11 [14]

Answer:

The length side of the pre-image is 16 units

Step-by-step explanation:

we know that

The length side of the image is equal to the length side of the pre-image multiplied by the scale factor

The length side of the pre-image is equal to the length side of the image divided by the scale factor

in this problem we have that

The scale factor is 1/2

The length side of the image is 8 units

therefore

8/(1/2)=16 units

The length side of the pre-image is 16 units

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2 years ago

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Given the image below, what is the image of C after a dilation with a scale factor of 3

Ne4ueva [31]

Answer: The coordinates of point C after the dilation are (-2, 5)

Step-by-step explanation:

I guess that you want to find where the point C ends after the dilation.

Ok, if we have a point (x, y) and we do a dilation with a scale A around the point (a,b), then the dilated point will be:

(a + A*(x - a), b + A*(y - b))

In this case we have:

(a,b) = (2,1) and A = 3.

And the coordinates of point C, before being dilated, are: (1, 2)

Then the new location of the point C will be:

C' = (1 + 3*(1 - 2), 2 + 3*(2 - 1)) = (1 -3, 2 + 3) = (-2, 5)

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2 years ago

Triangle J K L is shown. Angle J K L is a right angle. An altitude is drawn from point K to point M on side L J to form a right

kompoz [17]

Answer:

$LJ=15\ units$

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the length side KJ

In the right triangle JKM

Applying the Pythagoras Theorem

$KJ^{2}=JM^{2}+KM^{2}$

we have

$JM=3\ units$

$KM=6\ units$

substitute

$KJ^{2}=3^{2}+6^{2}$

$KJ^{2}=45}$

$KJ=\sqrt{45}\ units$

simplify

$KJ=3\sqrt{5}\ units$

step 2

Find the value of cosine of angle MJK in the right triangle JKM

$cos(JKM)=JM/KJ$

substitute the values

$cos(JKM)=\frac{3}{3\sqrt{5}}$

simplify

$cos(JKM)=\frac{\sqrt{5}}{5}$ -----> equation A

step 3

Find the value of cosine of angle MJK in the right triangle JKL

$cos(JKM)=KJ/LJ$

we have

$KJ=3\sqrt{5}\ units$

$cos(JKM)=\frac{\sqrt{5}}{5}$ ----> remember equation A

substitute the values

$\frac{\sqrt{5}}{5}=\frac{3\sqrt{5}}{LJ}$

Simplify

$LJ=5(3)=15\ units$

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2 years ago

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matthew is planning dinners for the next 3 nights. there’s are 11 meals to choose from. If no meal is repeated, how many differe

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Answer:

may be I am not sure 11

Step-by-step explanation:

i think ok just a guess

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Of 162 students honored at an academic awards banquet, 48 won awards for mathematics and 78 won awards for English. There are 14

ella [17]

Answer: 56/81

Step-by-step explanation:

in the attachment

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