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3 years ago

PLEASE ANSWER I REALLY NEED HELP ON THIS QUESTION!

Mathematics

1 answer:

SCORPION-xisa [38]3 years ago

8 0

The area of triangle BXA is 15*sqrt(2).

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Which is the best approximation for <img src="https://tex.z-dn.net/?f=%5Csqrt%7B90%7D" id="TexFormula1" title="\sqrt{90}" alt="\

Aleonysh [2.5K]

Answer:

A9.5

Step-by-step explanation:

I used a calculator......hope it helped:)))))))))))

4 0

3 years ago

Which is the completely factored form of 3xsquared -12x-15?

tekilochka [14]

Answer:

3(x - 5)(x + 1).

Step-by-step explanation:

3x^2 - 12x - 15

Dividing through by 3:

= 3(x^2 - 4x - 5)

We need 2 numbers whose sum is -4 and whose product = -5. That is -5 and +1 , so the factors are:

3(x - 5)(x + 1).

7 0

3 years ago

Can someone help me with this

melisa1 [442]

First part
option 2 p=KB

second part
p=$155,000k

or we can also find the value of k first
then the answer will be different :/

7 0

1 year ago

I think the answer is 38 degrees I just want someone to check if I am right please!

Aleksandr [31]

Answer:

JML = 142

Step-by-step explanation:

Sadly that is wrong. Not all the angles are the same. However, using the reflexive property we can find that angle JKM is 38 degrees. Using this, we need to create an equation to get the other two angles. Again, they would be the same because of the reflexive property.

38 + 38 + x + x = 360

The total needs to be 360 because the shape has 4 sides.

$76+x^{2} =360$

Now you beed to get the variable to one side. To do this you would subtract each side by 76.

$x^{2} =284$

Finally, your answer currently shows the sum of both LKJ and JML. to get one sides angle just divide by 2.

284 / 2 = 142

JML = 142

There is your answer.

Hope this helped.

4 0

2 years ago

What is the vertex form of Y = x2 - 6x + 6?

ahrayia [7]

Answer:

$\large\boxed{y=(x-3)^2-3}$

Step-by-step explanation:

The vertex form of a quadratic equation y = ax² + bx + c:

y = a(x - h)² + k

(h, k) - coordinates of a vertex

We have the equation y = x² - 6x + 6.

Convert to the vertex form:

$y=x^2-2(x)(3)+6\\\\y=\underbrace{x^2-2(x)(3)+3^2}_{(*)}-3^2+6\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\qquad(*)\\\\y=(x-3)^2-9+6\\\\y=(x-3)^2-3\to h=3,\ k=-3$

7 0

2 years ago