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pav-90 [236]
2 years ago
5

If (x + 3, 8) = (6, y – 2), then coordinates (x, y) issomeone help me :(​

Mathematics
1 answer:
DochEvi [55]2 years ago
7 0

Answer:

You can install ( Foto math) app then you can find it. Small help

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Martin recorded the low temperatures at his house for one week. The temperatures are shown below -7, -3, four, one, -2, -8, seve
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5,28-(x4-3,,,,,,,,,,,,,,,,,
7 0
3 years ago
The isosceles trapezoids, ABCD and EFGH, are similar quadrilaterals. The scale factor between the trapezoids is 2:3, = 6 centime
Bingel [31]

Answer:

Part A) see the explanation

Part B) see the explanation

Part C) The perimeter of trapezoid ABCD is 32 centimeters and the perimeter of trapezoid EFGH is 48 centimeters

Step-by-step explanation:

<u><em>The complete question is</em></u>

The isosceles trapezoids, ABCD and EFGH, are similar quadrilaterals. The scale factor between the trapezoids is 2:3, GH = 6 centimeters, AD = 8 centimeters, and AB is three times the length of DC

Part A) Write a similarity statement for each of the four pair of corresponding sides

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional

The corresponding sides are

AB and EF

BC and FG

CD and GH

AD and EH

so

\frac{AB}{EF}=\frac{BC}{FG}=\frac{CD}{GH}=\frac{AD}{EH}

Part B) Write a congruence statement for each of the four pair of corresponding angles.

we know that

If two figures are similar, then its corresponding angles are congruent

The corresponding angles are

∠A and ∠E

∠B and ∠F

∠C and ∠G

∠D and ∠H

so

∠A ≅ ∠E

∠B ≅ ∠F

∠C ≅ ∠G

∠D ≅ ∠H

Part C) Determine the perimeter for each of the isosceles trapezoids

we have

The scale factor between the trapezoids is 2:3, GH = 6 centimeters, AD = 8 centimeters, and AB is three times the length of DC

step 1

Find the measure of CD

Remember that

The ratio between corresponding sides is proportional and this ratio is the scale factor

Let

z ----> the scale factor

z=\frac{2}{3}

so

\frac{2}{3}=\frac{CD}{GH}}

we have

GH=6\ cm

substitute

\frac{2}{3}=\frac{CD}{6}}\\\\CD=4\ cm

step 2

Find the measure of AB

Remember that

AB is three times the length of DC

so

AB=3DC

DC=CD=4\ cm

substitute

AB=3(4)=12\ cm

step 3

Find the measure of BC

Remember that in an isosceles trapezoid, the legs are equal

so

BC=AD

we have

AD=8\ cm

therefore

BC=8\ cm

step 4

Find the perimeter of trapezoid ABCD

P=AB+BC+CD+AD

substitute the values

P=12+8+4+8=32\ cm

step 5

Find the perimeter of trapezoid EFGH

we know that

If two figures are similar, then the ratio of its perimeters is equal to the scale factor

Let

z ---> the scale factor

x ----> the perimeter of trapezoid ABCD

y ----> the perimeter of trapezoid EFGH

so

z=\frac{x}{y}

we have

z=\frac{2}{3}

x=32\ cm

substitute

\frac{2}{3}=\frac{32}{y}

y=32(3)/2=48\ cm

therefore

The perimeter of trapezoid EFGH is 48 centimeters

3 0
2 years ago
Monica wrote a 5 digit number 12,335 then circled the digit 3 on the right. which other digit in the number has a value that is
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6 0
3 years ago
1)Choose the graph below that correctly represents the equation 15x - 5y = 30.
Grace [21]
Pretty sure it is D
if you solve for the intercepts you get these answers
y-intercept is 6
x intercept is 2
7 0
3 years ago
Read 2 more answers
What is the equation of the quadratic graph with a focus of (-4,17/8) and a directrix of y=15/8
Temka [501]
So hmm notice the graph below

based on where the focus point is at, and the directrix, then, the parabola is opening upwards, meaning the squared variable is the "x"

\bf \begin{array}{llll}&#10;(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\&#10;\boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}})}\\&#10;\end{array}&#10;\qquad &#10;\begin{array}{llll}&#10;vertex\ ({{ h}},{{ k}})\\\\&#10;{{ p}}=\textit{distance from vertex to }\\&#10;\qquad \textit{ focus or directrix}&#10;\end{array}

now, keep in mind, the vertex is at coordinates h,k
the vertex itself is half-way between the focus and directrix
the directrix is at y=15/8 and the focus is at y=17/8
so, half-way will then be \bf \cfrac{17}{8}-\cfrac{15}{8}=\cfrac{2}{8}\iff \cfrac{1}{4}

well, so is 1/4 between the focus point and the directrix, half of that is 1/8
so, if you move from the focus point 1/8 down, you'll get the y-coordinate for the vertex, or 1/8 up from the directrix, since the vertex is equidistant to either

what's the "p" distance? well, we just found it, is just 1/8

so, the x-coordinate is obviously -4, get the y-coordinate by 17/8 - 1/8   or 15/8 + 1/8

and plug your values (x-h)² = 4p(y-k)   and then solve for "y", that's the equation of the quadratic

4 0
3 years ago
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