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2 years ago

Not sure how to solve this

Mathematics

1 answer:

inessss [21]2 years ago

8 0

Answer:

see below

Step-by-step explanation:

The x intercepts are where it crosses the x axis

( -4,0)

The y intercepts are where it crosses the y axis

( 0, 4)

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Seven less than a number, divided by 5, is -2

ycow [4]

Answer:

$\frac{x - 7}{5} = 2 \\cross \: multiply \\ x - 7 = 10 \\ x = 17$

hope this helps you

4 0

2 years ago

Which is equivalent to RootIndex 3 StartRoot 8 EndRoot Superscript one-fourth x?

Nat2105 [25]

<h3>Answer: Choice C</h3>

RootIndex 12 StartRoot 8 EndRoot Superscript x

12th root of 8^x = (12th root of 8)^x

$\sqrt[12]{8^{x}} = \left(\sqrt[12]{8}\right)^{x}$

=========================================

Explanation:

The general rule is

$\sqrt[n]{x} = x^{1/n}$

so any nth root is the same as having a fractional exponent 1/n.

Using that rule we can say the cube root of 8 is equivalent to 8^(1/3)

$\sqrt[3]{8} = 8^{1/3}$

-----

Raising this to the power of (1/4)x will have us multiply the exponents of 1/3 and (1/4)x like so

(1/3)*(1/4)x = (1/12)x

In other words,

$\left(8^{1/3}\right)^{(1/4)x} = 8^{(1/3)*(1/4)x}$

$\left(8^{1/3}\right)^{(1/4)x} = 8^{(1/12)x}$

-----

From here, we rewrite the fractional exponent 1/12 as a 12th root. which leads us to this

$8^{(1/12)x} = \sqrt[12]{8^{x}}$

$8^{(1/12)x} = \left(\sqrt[12]{8}\right)^{x}$

4 0

2 years ago

Read 2 more answers

What can you infer about the population from each data set represented below?

-BARSIC- [3]

The population seems like a young age group.

8 0

2 years ago

Please answer will see about brainliest

mina [271]

it is d. Pensacola beach

3 0

2 years ago

Solve these linear equations by Cramer's Rules Xj=det Bj / det A:

timurjin [86]

Answer:

(a) $x_1=-2,x_2=1$

(b) $x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}$

Step-by-step explanation:

(a) For using Cramer's rule you need to find matrix $A$ and the matrix $B_j$ for each variable. The matrix $A$ is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get $A$ more easily.

$2x_1+5x_2=1\\x_1+4x_2=2$

$\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]$

To get $B_1$ , replace in the matrix A the 1st column with the results of the equations:

$B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]$

To get $B_2$ , replace in the matrix A the 2nd column with the results of the equations:

$B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]$

Apply the rule to solve $x_1$ :

$x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2$

In the case of B2, the determinant is going to be zero. Instead of using the rule, substitute the values of the variable $x_1$ in one of the equations and solve for $x_2$ :

$2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1$

(b) In this system, follow the same steps,ust remember $B_3$ is formed by replacing the 3rd column of A with the results of the equations:

$2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0$

$\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]$

$B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]$

$B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]$

$B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]$

$x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}$

$x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}$

$x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}$

6 0

2 years ago