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3 years ago

13

A quadratic function is shown.

1 answer:

Yuki888 [10]3 years ago

4 0

The other factor would be x - 3

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Whats the best expansion for (x+5a+5b)(x+a-b)

jeka57 [31]

Answer:

x² + 6ax + 4bx + 5a² - 5b²

Step-by-step explanation:

(x+5a+5b )(x+a-b)

x² +a x -bx +5ax + 5a² -5ab + 5bx + 5ab -5b²

x² + 6ax + 4bx + 5a² - 5b²

4 0

3 years ago

I realllly neeedddd a aswerrr right now

ElenaW [278]

Answer:

for the first one it is "4/3"

for that second row i got "1"

for the third i got "12/13"

u dunno if you could do the same number twice but i got one

Step-by-step explanation:

terrible at explaining

5 0

3 years ago

Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro

strojnjashka [21]

Answer:

The missing side is $B = 6.0\ cm$

The missing angles are $\alpha = 56.2$ and $\theta = 93.8$

Step-by-step explanation:

Given

$A = 10\ cm$

$C = 12\ cm$

$\beta = 30$

The implication of this question is to solve for the missing side and the two missing angles

Represent

Angle A with $\alpha$

Angle B with $\beta$

Angle C with $\theta$

Calculating B

This will be calculated using cosine formula as thus;

$B^2 = A^2 + C^2 - 2ACCos\beta$

Substitute values for A, C and $\beta$

$B^2 = 10^2 + 12^2 - 2 * 10 * 12 * Cos30$

$B^2 = 100 + 144 - 240 * 0.8660$

$B^2 = 100 + 144 - 207.8$

$B^2 = 36.2$

Take Square root of both sides

$B = \sqrt{36.2}$

$B = 6.0$ <em>(Approximated)</em>

Calculating $\alpha$

This will be calculated using cosine formula as thus;

$A^2 = B^2 + C^2 - 2BCCos\alpha$

Substitute values for A, B and C

$A^2 = B^2 + C^2 - 2BCCos\alpha$

$10^2 = 6^2 + 12^2 - 2 * 6 * 12 * Cos\alpha$

$100 = 36 + 144 - 144Cos\alpha$

Collect Like Terms

$100 - 36 - 144 = -144Cos\alpha$

$-80 = -144Cos\alpha$

Divide both sides by -144

$\frac{-80}{-144} = Cos\alpha$

$0.5556 = Cos\alpha$

$\alpha = cos^{-1}(0.5556)$

$\alpha = 56.2$ <em>(Approximated)</em>

Calculating $\theta$

This will be calculated using cosine formula as thus;

$C^2 = B^2 + A^2 - 2BACos\theta$

Substitute values for A, B and C

$12^2 = 6^2 + 10^2 - 2 * 6 * 10Cos\theta$

$144 = 36 + 100 - 120Cos\theta$

Collect Like Terms

$144 - 36 - 100 = -120Cos\theta$

$8 = -120Cos\theta$

Divide both sides by -120

$\frac{8}{-120} = Cos\theta$

$-0.0667= Cos\theta$

$\theta = cos^{-1}(-0.0667)$

$\theta = 93.8$ <em>(Approximated)</em>

4 0

3 years ago

Using the digits 2,3,5,6,8 find the smallest possible product?

MA_775_DIABLO [31]

the product is always the same:

1440

5 0

3 years ago

Help plz and thank u ;D

igomit [66]

Put this sign in the blank >

3 0

4 years ago

Read 2 more answers