All categories

3 years ago

Help please this is an important grade

Mathematics

1 answer:

faust18 [17]3 years ago

6 0

Answer:

it might be c or b

Step-by-step explanation:

You might be interested in

The radius of a circle is 13 miles. What is the circle's area?

lesantik [10]

A=π×r×2=

=π×13×2 =

530.92916 mi²

here is your answer

hope it will help u

3 0

2 years ago

Read 2 more answers

A diagonal of a cube measures

Gennadij [26K]

Answer: 7cm

Step-by-step explanation:

Use pythagorean theorum:

Variables:

Let the edge = A

Face diagonal (F) = 10cm

Inner diagonal (I) = sqrt(150) cm

A^2 = I^2 - F^2

A^2 = [sqrt(150)]^2 - 10^2

A^2 = 150 - 100

A^2 = 50

A = sqrt(50)

A = 7.0710678 cm

A = 7cm (to the nearest tenth)

so, length of an edge = 7cm

4 0

2 years ago

3(-21)/7-12/6 need the answer

Mademuasel [1]

If ur equation looks like this, then the answer is -63/5 , and in decimal form the answer is -12.6 i hope this helps! !!

7 0

3 years ago

Is the quotient of (-5) ÷ (-7) a rational number

Anarel [89]

Yes it is. 5\7 is a fraction, therefore it is rational; it can be written as a decimal, 0.714285...... [bar notation indication].

4 0

3 years ago

Read 2 more answers

Write the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10).

Serjik [45]

Answer:

$\displaystyle f(x)=x^2+2x+2$

Step-by-step explanation:

<u>System Of Linear Equations </u>

In this problem, we'll need to solve a 3x3 system of linear equations because we have three unknowns and three conditions.

We are required to find the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10)

The general quadratic function can be written as

$\displaystyle f(x)=ax^2+bx+c$

We need to find the values of a,b, and c. Let's use the first condition, i.e. f(-1)=1

$\displaystyle f(-1)=a(-1)^2+b(-1)+c$

$\displaystyle f(-1)=a-b+c$

$\displaystyle a-b+c=1.....[eq\ 1]$

Now we use the second condition f(1)=5

$\displaystyle f(1)=a(1)^2+b(1)+c$

$\displaystyle f(1)=a+b+c$

$\displaystyle a+b+c=5.......[eq\ 2]$

Finally, we use the third condition f(2)=10

$\displaystyle f(2)=a(2)^2+b(2)+c$

$\displaystyle f(2)=4a+2b+c$

$\displaystyle 4a+2b+c=10....[eq\ 3]$

We put together eq 1, eq 2, and eq 3 to form the system

$\displaystyle \left\{\begin{matrix}a-b+c=1\\ a+b+c=5\\ 4a+2b+c=10\end{matrix}\right.$

Adding the first two equations we have

$\displaystyle 2a+2c=6$

$\displaystyle a+c=3$

And also

$\displaystyle b=2$

Using the above equation and the value of b in the third equation, we have

$\displaystyle \left\{\begin{matrix}a+c=3\\ 4a+c=6\end{matrix}\right.$

Subtracting the first equation from the second

$\displaystyle 3a=3$

$\displaystyle a=1$

And therefore

$\displaystyle c=2$

Now we have all the values, the quadratic function is

$\displaystyle \boxed{f(x)=x^2+2x+2}$

6 0

3 years ago