Ask question

Ask question

All categories

3 years ago

10

Write the expression (4x-2).(6x 7) in the standard form of a anquadratic expression,ax^2 bx c what are the values of the coeffic

ient of each term and the constant term

1 answer:

kipiarov [429]3 years ago

6 0

Hello:
<span>(4x-2)(6x+ 7)=24x²+28x-12x-14
= 24x²+16x-14

</span><span>in the standard form : ax²+bx+c a = 24 b=16 c=-14</span>

You might be interested in

Need help due in 10 mins

Marina CMI [18]

Answer:

A) The amount of money deposited monthly

Step-by-step explanation:

y = 35x (slope) (amount deposited each month)+ 250 (y-intercept) (starting amount)

4 0

2 years ago

What is the answer for this -12n - (-13n)=

Ksenya-84 [330]

Answer:

n

Step-by-step explanation:

-12n - (-13n) = -12n +13n = 1n

8 0

3 years ago

Factor the algebraic expression. 40a+5

Vitek1552 [10]

Answer:

5 (8a + 1)

Step-by-step explanation:

Since they both can be factored by 5 you would divide both by 5 and get 8 and 1.

7 0

3 years ago

55.405 greater than 55.045

Kobotan [32]

55.405 is bigger then 55.045

4 0

3 years ago

Read 2 more answers

Find the coordinates of the three points on the graph of y = (1/2)x² -2x whose x-values are −2, 0, and 4.

MaRussiya [10]

Answer:

y = 6

y = 0

y = 0

Step-by-step explanation:

$y=\frac{1}{2}x² -2x$

When, x = -2

$y=\frac{1}{2}\times (-2)^2-2\times(-2)\\\Rightarrow y=\frac{1}{2}4+4\\\Rightarrow y=2+4\\\Rightarrow y=6$

y = 6

When, x = 0

$y=\frac{1}{2}\times (0)^2-2\times(0)\\\Rightarrow y=0+0\\\Rightarrow y=0$

y = 0

When, x = 4

$y=\frac{1}{2}\times (4)^2-2\times(4)\\\Rightarrow y=\frac{1}{2}16-8\\\Rightarrow y=8-8\\\Rightarrow y=0$

y = 0

3 0

3 years ago