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3 years ago

Can someone please answer number 7. The first question. Pleaseeeeee

Mathematics

2 answers:

ruslelena [56]3 years ago

8 0

Answer:

(3) {5,7}

Step-by-step explanation

135 people voted for it on gauth math

Helen [10]3 years ago

3 0

Answer:

Step-by-step explanation:

5^2-4(5)-5=2(5)-10

0=0

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What is the following sum in simplest form - see attached.

melomori [17]

You must first simplify the radicals by taking out any perfect squares.

√8 + 3√2 + √32 =
√4 * √2 + 3√2 + √16 * √2 =
2√2 + 3√2 + 4√2 =
9√2

5 0

3 years ago

Y=x^2-6x-16 in vertex form

satela [25.4K]

Answer:

$y=(x-3)^{2} -25$

Step-by-step explanation:

The standard form of a quadratic equation is $y=ax^{2} +bx+c$

The vertex form of a quadratic equation is $y=a(x-h)^{2} +k$

The vertex of a quadratic is (h,k) which is the maximum or minimum of a quadratic equation. To find the vertex of a quadratic, you can either graph the function and find the vertex, or you can find it algebraically.

To find the h-value of the vertex, you use the following equation:

$h=\frac{-b}{2a}$

In this case, our quadratic equation is $y=x^{2} -6x-16$ . Our a-value is 1, our b-value is -6, and our c-value is -16. We will only be using the a and b values. To find the h-value, we will plug in these values into the equation shown below.

$h=\frac{-b}{2a}$ ⇒ $h=\frac{-(-6)}{2(1)}=\frac{6}{2} =3$

Now, that we found our h-value, we need to find our k-value. To find the k-value, you plug in the h-value we found into the given quadratic equation which in this case is $y=x^{2} -6x-16$

$y=x^{2} -6x-16$ ⇒ $y=(3)^{2} -6(3)-16$ ⇒ $y=9-18-16$ ⇒ $y=-25$

This y-value that we just found is our k-value.

Next, we are going to set up our equation in vertex form. As a reminder, vertex form is: $y=a(x-h)^{2} +k$

a: 1

h: 3

k: -25

$y=(x-3)^{2} -25$

Hope this helps!

3 0

4 years ago

The equation of a circle is (x + 4)2 + (y + 6)2 = 16. Determine the length of the radius.

faust18 [17]

R^2=16
Therefore R=4

7 0

3 years ago

Read 2 more answers

Pls help me solve this

melamori03 [73]

Answer:

See below.

Step-by-step explanation:

1.) False - It contains 6 sig figs, not 2.

2.) False - It goes up to 2 decimal places, not 3.

3.) True

4.) False - It goes up to 1 sig fig, not 4.

8 0

3 years ago

A fraction becomes 4÷5 if 1 is added to both numerator and denominator. If, however, 5 is subtracted from both numerator and den

tia_tia [17]

Answer:

$\dfrac{7}{9}$

Step-by-step explanation:

$\dfrac{x+1}{y+1}=\dfrac{4}{5}\\\Rightarrow 5x-4y=-1$

$\dfrac{x-5}{y-5}=\dfrac{1}{2}\\\Rightarrow 2x-y=5$

Putting it in matrix form

$\begin{bmatrix}a_{1}&b_{1}\\a_{2}&b_{2}\end{bmatrix}{\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}{c_{1}}\\{c_{2}}\end{bmatrix}\\\Rightarrow\begin{bmatrix}5 & -4\\2 & -1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}-1\\ 5\end{bmatrix}$

From Cramer's rule we have

$x=\dfrac{\begin{vmatrix}c_1 &b_1 \\ c_2 & b_2\end{vmatrix}}{\begin{vmatrix}a_1 &b_1 \\ a_2& b_2\end{vmatrix}}\\\Rightarrow x=\dfrac{\begin{vmatrix}-1 &-4 \\ 5 & -1\end{vmatrix}}{\begin{vmatrix}5 &-4 \\ 2& -1\end{vmatrix}}\\\Rightarrow x=\dfrac{1+20}{-5+8}\\\Rightarrow x=7$

$y=\dfrac{\begin{vmatrix}a_1 &c_1 \\ a_2 & c_1\end{vmatrix}}{\begin{vmatrix}a_1 &b_1 \\ a_2& b_2\end{vmatrix}}\\\Rightarrow y=\dfrac{\begin{vmatrix}5 &-1 \\ 2 & 5\end{vmatrix}}{\begin{vmatrix}5 &-4 \\ 2& -1 \end{vmatrix}}\\\Rightarrow y=\dfrac{25+2}{-5+8}\\\Rightarrow y=9$

Verifying the results

$\dfrac{7+1}{9+1}=\dfrac{8}{10}=\dfrac{4}{5}$

$\dfrac{7-5}{9-5}=\dfrac{2}{4}=\dfrac{1}{2}$

Hence, the fraction is $\dfrac{7}{9}$ .

6 0

3 years ago