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3 years ago

Help !!!!! GEOMETRY PEOPLE HELP ME LOVES

Mathematics

2 answers:

chubhunter [2.5K]3 years ago

4 0

The new triangle produced is exactly identical to the parent triangle - no change in size, shape, rotation, or vertex position. This transformation is, therefore, an isometry.

Brrunno [24]3 years ago

3 0

The new triangle produced is an exact remake of the first one. The angles or size does not change between the two. Thus, this transformation is an Isometry.

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Guys please help me

stich3 [128]

Answer: (x + [-1], y + [1])

Step-by-step explanation:

<em>See attached. </em>We can draw, or picture it in our heads, what the reflection would look like. Then we can pick one (or multiple to test) points and see the translation.

We can also test with a set of points. B', (2, 4) becomes G in the transformation. G is at (1, 5)

(1 - 2, 5 - 4) -> (-1, 1)

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1 year ago

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What is 64 thousandths in decimal form

shusha [124]

I am pretty sure it is 0.064. Hope this helps

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3 years ago

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-3a+2b=-3<br> solve for a

vivado [14]

-3a=-3-2b
a=1+2/3b

Solution
a=1+2/3b

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3 years ago

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Whoever checks my work and tell what I did wrong and the right answer will

gulaghasi [49]

Answer:

See below

Step-by-step explanation:

I believe that you only had to do letters F, H, and J. In that case, let's go over each one!

F: For isolating $d_{1}$ , we need to get rid of the 1/2 first. Let's multiply each side by 2:

$2*m = 2*\frac{1}{2}(d_{1}+ d_{2}) \\2m = d_{1}+ d_{2}$

After this, we just subtract $d_{2}$ from each side to get $d_{1}=2m- d_{2}$ . Dark Blue is correct! Let's now plug in those numbers below:

$d_{1}=2(10)- 13 = 20-13=7$

G: Let's isolate the vw^2 on one side by subtracting y from each side:

$vw^{2} +y-y=x-y\\vw^{2}=x-y$

Let's now divide each side by v, then put each side under a square root to get our final answer:

$\frac{vw^2}{v} = \frac{x-y}{v}\\ w^2 = \frac{x-y}{v}\\\sqrt{w^2} = \sqrt{\frac{x-y}{v}} \\w=\sqrt{\frac{x-y}{v}}$

Orange is correct! Again, let's solve the problem underneath:

w= $w=\sqrt{\frac{38-(-7)}{5}} = \sqrt{\frac{38+7}{5}} = \sqrt{\frac{45}{5}}=\sqrt{9}=3$

H: This one has some stuff that we haven't worked with quite yet (like terms), but our approach is the same: isolate c on one side of the equation.

$2a-2a+3c=17a-2a+21\\3c = 15a+21\\\frac{3c}{3} = \frac{15a+21}{3}\\c = \frac{15a}{3}+ \frac{21}{3} \\c = 5a+7$

Purple is correct! Let's solve the problem:

$c = 5(\frac{16}{5})+7 = 16+7 = 23$

7 0

3 years ago

Please solve this sheet.

lyudmila [28]

Hello,

A: roots: -1,-3
a point (-2,1)
Vertex=((-2,1)

y=k*(x+1)(x+3) using roots
but k*(-2+1)(-2+3)=1==>k*(-1)*1=1==>k=-1

eq: y=-(x+1)(x+3)
==>y=-(x²+3x+x+3)
==>y=-x²-4x-3
y=k(x+2)²+1 if x=-1,y=0 ==>k*1+1=0==>k=-1
==>y=-(x+2)²+1

Answer :A--> R,K

B)
y=k(x+4)²-2 and k=-1/2
y=-1/2(x+4)²-2
y=-1/2x²-4x-10

answer B--> I,≈W if it is written -1/2*x² (square has been forgotten)

C:
y=2x²-16x+30
y=2(x-4)²-2
answer : C-->S,J

D:
y=-(x+3)(x+1)
y=-x²-4x-3
=-(x+2)²+1
answer D--> V,L

E:
Here there is a problem: or the graph is wrong, or 2 equations are missing!

y=1(x+1)(x-3) using roots
y=x²-2x-3 ≈ T si it were -2x and not +2x.

y=(x-1)²-4 ≈H is it were -1 in place of +1 [H:y=(x+1)²-4]

6 0

3 years ago