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2 years ago

How is the graph of y = 3x^2 + 7 different from the graph of y = 3x^2?

Mathematics

1 answer:

Lana71 [14]2 years ago

4 0

The 1st one is not linear but the second one is.

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If 14 out of 20 is correct what percent is that?

Kobotan [32]

If 14 out of 20 is correct, then that is : 14/20 = 0.70 = 70%

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3 years ago

Read 2 more answers

Gia opened two savings accounts at two diffrent banks. One account earns an annual 3.4% simple interest, and the other earns hal

dsp73

Interest from first account is :

$I_1 = \dfrac{p\times r \times t}{100}\\\\I_1 = \dfrac{509\times 3.4\times 3}{100}\\\\I_1 =\$ 51.918$

Interest form second account is :

$I_2 = \dfrac{p\times r\times t}{100}\\\\I_2 = \dfrac{509\times 1.7\times 3}{100}\\\\I_2 = \$25.959$

So, total interest is :

$I = I_1 + I_2 \\\\I = 51.918 + 25.959\\\\I = \$77.877$

Therefore, total interest is $77.877 .

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3 years ago

A line and the point (3,4) are graphed in the coordinate plane. what is the equation of the line that passes through the point (

oksano4ka [1.4K]

Answer:

B. y=1/4x+13/4

Step-by-step explanation:

3 0

2 years ago

How to solve Which ordered pair is a solution for x − 3y = 1?

babunello [35]

There is no one answer (ordered pair) to this equation and the answer is found by graphing or trying out different values for each variable. Anyway, here are some solutions; (1,0), (4,1), (0,-1/3).

5 0

3 years ago

To two decimal places, find the value of k that will make the function f(x) continuous everywhere. f of x equals the quantity 3x

Fiesta28 [93]

Given function is

$f(x)=\left\{\begin{matrix}3x+k & x\leq -4 \\ kx^2-5 & x> -4\end{matrix}\right.$

now we need to find the value of k such that function f(x) continuous everywhere.

We know that any function f(x) is continuous at point x=a if left hand limit and right hand limits at the point x=a are equal.

So we just need to find both left and right hand limits then set equal to each other to find the value of k

To find the left hand limit (LHD) we plug x=-4 into 3x+k

so LHD= 3(-4)+k

To find the Right hand limit (RHD) we plug x=-4 into

$kx^2-5$

so RHD= $k(-4)^2-5$

Now set both equal

$k(-4)^2-5=3(-4)+k$

$16k-5=-12+k$

$16k-k=-12+5$

$15k=-7$

$k=-\frac{7}{15}$

k=-0.47

<u>Hence final answer is -0.47.</u>

7 0

3 years ago