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3 years ago

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Please help guys! please please help

1 answer:

gtnhenbr [62]3 years ago

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What is the value of this expression when a = 4, b= -5, and c= -7?<br> a + 2bc<br> 3a ?

sladkih [1.3K]

Answer:

1) -66

2)

Step-by-step explanation:

Given:

a=4

b=-5

c=-7

=a+2bc

=4+2*-5*7

=4+(-10*7)

=4-70

=-66

=3a

=3(4)

=12

Hope this helps ;) ❤❤❤

7 0

3 years ago

Read 2 more answers

triangle abc is rotated 90° clockwise about the origin. what is the rules that we can use to find the vertices location of the i

yuradex [85]

Answer:

The answer is when we rotating something 90 degrees clockwise about the orgin

(x,y) goes to (y,-x)

5 0

2 years ago

18.4b=44.16 need help fast

MaRussiya [10]

Answer:

<u>If it's 18.4b=44.16:</u> $b=2.4$

<u>If it's 4b=44.16:</u> $b=11.04$

Step-by-step explanation:

<u>If it's 18.4b=44.16:</u>

$18.4b=44.16\\\\\frac{b}{18.4}=\frac{44.16}{18.4}\\\\b=2.4$

<u />

<u>If it's 4b=44.16:</u>

<u /> $4b=44.16\\\\\frac{b}{4}=\frac{44.16}{4}\\\\b=11.04$ <u />

4 0

3 years ago

The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

andrezito [222]

Answer:

Given the equation: $3x^2+10x+c =0$

A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

$x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}$

Simplify:

$x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}$

Also, it is given that the difference of two roots of the given equation is $4\frac{2}{3} = \frac{14}{3}$

i.e,

$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

$x_2= \frac{-10 - \sqrt{100- 12c}}{6}$ .....[3]

then;

$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

simplify:

$\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}$

or

$\sqrt{100- 12c} = 14$

Squaring both sides we get;

$100-12c = 196$

Subtract 100 from both sides, we get

$100-12c -100= 196-100$

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve $x_1 , x_2$

$x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}$

or

$x_1 = \frac{-10 + \sqrt{100- 96}}{6}$ or

$x_1 = \frac{-10 + \sqrt{4}}{6}$

Simplify:

$x_1 = \frac{-4}{3}$

Now, to solve for $x_2$ ;

$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

or

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

Simplify:

$x_2 = -2$

therefore, the solution for the given equation is: $-\frac{4}{3}$ and -2.

3 0

3 years ago

HELP ME........................................................................

GaryK [48]

Answer: 17
2^2 is 4
45/9 is 5
Multiply 4•5 to get 20
Subtract 3

8 0

3 years ago

Read 2 more answers