All categories

3 years ago

May somebody help me please.

Mathematics

1 answer:

DiKsa [7]3 years ago

5 0

Answer:

y = 2 -> (-5, 2)

Step-by-step explanation:

To solve questions like these, you plug in the variable you know, which in this case, is x = -5.

That gives us -5 - 5y = -15.

Then, you solve it like a normal equation.

-5y = -10 (add 5 to both sides)

-y = -2 (divide by 5)

y = 2. (divide by -1)

You might be interested in

A small radio transmitter broadcasts in a 35 mile radius. if you drive along a straight line from a city 46 miles north of the t

gtnhenbr [62]

30 miles also yes you will hear static

3 0

4 years ago

8 circles and 20 squares what is the simplest ratio of total shapes

viva [34]

Answer

8:28 Hope this helps.

7 0

1 year ago

Prove the identity (sin6x)/(1+cos6x)=tan3x.

Vikentia [17]

Use double angle identity
$sin(2x) = 2 sin x cos x \\ \\ cos(2x) = 2cos^2 x - 1$

We can relate '6x' to '3x' in the same way since 6 is 2*3.
$sin(6x) = 2 sin (3x) cos (3x) \\ \\ cos(6x) = 2cos^2 (3x) - 1$

Now sub into left side of identity. Simplify until it equals right side.
$\frac{2 sin(3x) cos(3x)}{1+(2cos^2 (3x) -1)} = \frac{2 sin(3x) cos(3x)}{2cos^2 (3x) } = \frac{sin(3x)}{cos(3x)} = tan (3x)$

3 0

4 years ago

What is the volume of the prism that can be constructed from this net?

Elden [556K]

The aswer is A hop it helps

5 0

4 years ago

What is the equation of this line in standard form?

Fiesta28 [93]

keeping in mind that standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

$\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{\frac{1}{2}}~,~\stackrel{y_2}{2}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{(-1)}}}{\underset{run} {\underset{x_2}{\frac{1}{2}}-\underset{x_1}{(-3)}}}\implies \cfrac{2+1}{\frac{1}{2}+3}\implies \cfrac{3}{\frac{1+6}{2}}\implies \cfrac{\frac{3}{1}}{~~\frac{7}{2}~~}\implies \cfrac{3}{1}\cdot \cfrac{2}{7}\implies \cfrac{6}{7}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-1)}=\stackrel{m}{\cfrac{6}{7}}[x-\stackrel{x_1}{(-3)}]\implies y+1=\cfrac{6}{7}(x+3) \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{7}}{7(y+1)=7\left( \cfrac{6}{7}(x+3) \right)}\implies 7y+7=6(x+3)\implies 7y+7=6x+18 \\\\\\ 7y=6x+11\implies -6x+7y=11\implies 6x-7y=-11$

3 0

3 years ago