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bekas [8.4K]
3 years ago
11

Multiply.

Mathematics
1 answer:
Reptile [31]3 years ago
7 0

Answer:

correct answer is D

Step-by-step explanation:

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За(2b + 5c)<br><br> I need help
ch4aika [34]
Hey! Your answer would be = 6ab+15ac
8 0
3 years ago
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lilavasa [31]
I hope you know copying and pasting done things makes it a lil hard to read
7 0
4 years ago
A,B and C are the vertices of one triangle.
dsp73

Answer:


Step-by-step explanation:

Given: The triangle with coordinate A(4,6), B(2,-2) and C(-2,-4). D is the mid point of AB and E is the mid point of AC.

To prove: DE is parallel to BC.

Construction: Join DE.

Proof: If we prove the basic proportionality theorem that is \frac{AD}{DB}=\frac{AE}{EC}, then it proves that DE is parallel to BC.

Now, Mid Point D has coordinates=(\frac{4+2}{2},\frac{6-2}{2})=(3,2) and Mid Point E has coordinates=(\frac{4-2}{2},\frac{6-4}{2})=(1,1)

Now, AD= \sqrt{(4-3)^{2}+(6-2)^{2}}=\sqrt{17}

DB=\sqrt{(3-2)^{2}+(2+2)^{2}}=\sqrt{17}

AE=\sqrt{(4-1)^{2}+(6-1)^{2}}=\sqrt{34}

EC=\sqrt{(1+2)^{2}+(1+4)^{2}}=\sqrt{34}

Now, \frac{AD}{DB}=\frac{AE}{EC}

=\frac{\sqrt{17}}{\sqrt{17}}=\frac{\sqrt{34}}{\sqrt{34}}=\frac{1}{1}

Hence, \frac{AD}{DB}=\frac{AE}{EC}

Thus, By basic proportionality theorem, DE is parallel to BC.

4 0
4 years ago
Read 2 more answers
Find the standard form of the equation of the hyperbola satisfying the given conditions: X intercept +/- 6; foci at (-10,0) and
uysha [10]

Answer:

\frac{x^{2}}{36} - \frac{y^{2}}{64}=1

Step-by-step explanation:

Given an hyperbola with the following conditions:

  • Foci at (-10,0) and (10,0)
  • x-intercept +/- 6;

The following holds:

  • The center is midway between the foci, so the center must be at (h, k) = (0, 0).
  • The foci are 10 units to either side of the center, so c = 10 and c^2 = 100
  • The center lies on the origin, so the two x-intercepts must then also be the hyperbola's vertices.

Since the intercepts are 6 units to either side of the center, then a = 6 and a^2 = 36.

Then, a^2+b^2=c^2\\b^2=100-36=64

Therefore, substituting a^2 = 36. and b^2=64 into the standard form

\frac{x^{2}}{a^2} - \frac{y^{2}}{b^2}=1\\We \: have:\\ \dfrac{x^{2}}{36} - \dfrac{y^{2}}{64}=1

4 0
3 years ago
What step is incorrect
tatyana61 [14]

Answer:

3

Have a great day!!

5 0
2 years ago
Read 2 more answers
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