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3 years ago

8

How do i factor -4c^2+19c+5

1 answer:

Liula [17]3 years ago

3 0

Answer:

-1(4c+1)(c-5)

Step-by-step explanation:

First get rid of the negative by factoring out -1, then determine what factors of 5, the constant term(the number without the x), multiply by 4 to add up to -19. The only factors of 5 are 1 and 5, so we will use these. (4c+1)(c-5) can be "foiled" to check our answer.

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lakkis [162]

Answer:

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Step-by-step explanation:

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3 years ago

A store pays $85 for a picnic table and marks the price up by 25%. What is the new price

Setler79 [48]

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3 years ago

Read 2 more answers

Help ASAP! And explain!

stich3 [128]

Answer:

Option (2)

Step-by-step explanation:

Given:

AC is an angle bisector of ∠DAB and ∠DAB

m∠BCA ≅ m∠DCA

m∠BAC ≅ m∠DAC

To Prove:

ΔABC ≅ ΔADC

Solution:

Statements Reasons

1). m∠BCA ≅ m∠DCA 1). Given

2). m∠BAC ≅ m∠DAC 2). Given

3). AC ≅ AC 3). Reflexive property

4). ΔABC ≅ ΔADC 4). ASA property of congruence

Therefore, Option (2) will be the correct option.

4 0

3 years ago

Expand the expression <br><br><br> -7(k - 3)

Illusion [34]

How can I help you?
−
7

(
k
−
3
)
-
7

(
k
-
3
)
Solve

4 0

3 years ago

Section 5.2 Problem 6:<br><br>Find the general solution<br><img src="https://tex.z-dn.net/?f=y%27%27%20%2B%206y%27%20%2B%2010y%2

mihalych1998 [28]

Answer:

$y=e^{-3t}(A\: cos\: t+B\:sin\:t)$

Step-by-step explanation:

<u>Given Second-Order Homogenous Differential Equation</u>

$y''+6y'+10y=0$

<u>Use Auxiliary Equation</u>

<u /> $m^2+6m+10=0\\\\(m+3)^2+1=0\\\\(m+3)^2=-1\\\\m+3=\pm i\\\\m=-3\pm i$

<u>General Solution</u>

<u /> $y=e^{pt}(A\: cos\: qt+B\:sin\:qt)\\\\y=e^{-3t}(A\: cos\: t+B\:sin\:t)$

Note that the DE has two distinct complex solutions $p\pm qi$ where $A$ and $B$ are arbitrary constants.

4 0

2 years ago