Ask question

Ask question

All categories

3 years ago

7

Just 13 a and b please and thank you

1 answer:

ehidna [41]3 years ago

5 0

A)1.625m and b I can't read

You might be interested in

Write an addition sentence that helps you solve 15-9.

Lemur [1.5K]

Answer:

I have 15 cookies and my brother ate 9 now I have 6 cookies.

8 0

3 years ago

9 out of 25 students are home sick with the flu. What percentage of the students are home sick?

Arturiano [62]

Answer:

36%

Step-by-step explanation:

9/25

multiply each by 4 to get 36/100

as a percent it would be 36%

5 0

4 years ago

Read 2 more answers

Graph the Rarabola.---Plot five points on the parabola: the vertex, two points to the left of the vertex, and tvbutton2Explanati

valentinak56 [21]

Which parabola?

$y=-x^2$

Equation of the parabola

y - y1 = 4p(x - x1)

The Vertex of the parabola given is (0, 0) because it does not have the values of x1 and y1.

Then Vertex = (0,0)

-Look for two values of y to the left and two points to the right.

You can choose the points that you which

x y

-3 y = -(-3)^2 = -9

-1 y = -(-1)^2 = -1

0 y = -(0)^2 = 0

1 y = -(1)^2 = -1

3 y = -(3)^2 = -9

3 0

1 year 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

James puts $3,500 into a savings account that earns 2.5% simple interest. He does not touch that account for 3 years. What would

miskamm [114]

Given :

James puts $3,500 into a savings account that earns 2.5% simple interest.

He does not touch that account for 3 years.

To Find :

The new balance of the account be after 3 years.

Solution :

Interest on $3500 after 3 years is :

$I = \dfrac{P_o\times r\times t}{100}\\\\I = \dfrac{3500\times 2.5\times 3}{100}\\\\I = \$262.5$

So, new balance of the account after 3 years is $( 3500+262.5 ) = $3762.5 .

Hence, this is the required solution.

3 0

3 years ago