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2 years ago

Which equations are equivalent to y = two-thirds x minus 4 when written in slope-intercept form? Check all that apply.

Mathematics

2 answers:

Nikitich [7]2 years ago

5 0

Answer:

B an C is the answer

Step-by-step explanation:

I HOPE THIS HELPS!

I TOOK THE QUIZ ON Edge

mixas84 [53]2 years ago

5 0

Answer:

B) & C).

Step-by-step explanation:

Free points UuU.

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The general solution of 2 y ln(x)y' = (y^2 + 4)/x is

Sav [38]

Replace $y'$ with $\dfrac{\mathrm dy}{\mathrm dx}$ to see that this ODE is separable:

$2y\ln x\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y^2+4}x\implies\dfrac{2y}{y^2+4}\,\mathrm dy=\dfrac{\mathrm dx}{x\ln x}$

Integrate both sides; on the left, set $u=y^2+4$ so that $\mathrm du=2y\,\mathrm dy$ ; on the right, set $v=\ln x$ so that $\mathrm dv=\dfrac{\mathrm dx}x$ . Then

$\displaystyle\int\frac{2y}{y^2+4}\,\mathrm dy=\int\dfrac{\mathrm dx}{x\ln x}\iff\int\frac{\mathrm du}u=\int\dfrac{\mathrm dv}v$

$\implies\ln|u|=\ln|v|+C$

$\implies\ln(y^2+4)=\ln|\ln x|+C$

$\implies y^2+4=e^{\ln|\ln x|+C}$

$\implies y^2=C|\ln x|-4$

$\implies y=\pm\sqrt{C|\ln x|-4}$

4 0

2 years ago

Find the area of the right triangle.

zzz [600]

48 just count the inside of the triangle which is difficult since all side aren’t a full square or count the outside border of the triangle and just multiply I did 6x8

6 0

3 years ago

Read 2 more answers

How can I find the exact distance between the points(√7,-√2)and (4√7,5√2)

chubhunter [2.5K]

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ \sqrt{7}}}\quad ,&{{ -\sqrt{2}}})\quad 
% (c,d)
&({{4\sqrt{7}}}\quad ,&{{ 5\sqrt{2}}})
\end{array}~~~
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
$

$\bf d=\sqrt{[4\sqrt{7}-\sqrt{7}]^2~+~[5\sqrt{2}-(-\sqrt{2})]^2}
\\\\\\
d=\sqrt{(4\sqrt{7}-\sqrt{7})^2~+~(5\sqrt{2}+\sqrt{2})^2}\implies d=\sqrt{(3\sqrt{7})^2~+~(6\sqrt{2})^2}
\\\\\\
d=\sqrt{3^2\cdot 7~~+~~6^2\cdot 2}\implies d=\sqrt{63+72}\implies d=\sqrt{135}
\\\\\\
\begin{cases}
135=5\cdot 3\cdot 3\cdot 3\\
\qquad 5\cdot 3^2\cdot 3\\
\qquad 15\cdot 3^2
\end{cases}\implies d=\sqrt{15\cdot 3^2}\implies d=3\sqrt{15}$

8 0

3 years ago

there is a beaker of 3.5% acid solution and a beaker of 6% acid solution in the science lab. Mr. Larson needs 200ml of 4.5% acid

disa [49]

Below are the choices:

A. 80 mL of the 3.5% solution and 120 mL of the 6% solution

B. 120 mL of the 3.5% solution and 80 mL of the 6% solution 

C. 140 mL of the 3.5% solution and 60 mL of the 6% solution 

D. 120 mL of the 3.5% solution and 80 mL of the 6% solution

Let fraction of 3.5% in final solution be p.

p * 3.5 + (1 - p) * 6 = 4.5 

3.5p + 6 - 6p = 4.5 

2.5p = 1.5 

p = 3/5 

3/5 * 200 = 120 

Therefore the answer is B. 120 ml of 3.5% and 80 ml of 6%.

6 0

3 years ago

X/2.1 = -1.8 Show your work and check it (write it out plz)

Lerok [7]

X = -1.8 * 2.1
x = 3.78

don't know how i can write it longer

7 0

3 years ago