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3 years ago

Please help me I really need help

Mathematics

2 answers:

elixir [45]3 years ago

8 0

Answer:

Step-by-step explanation:

Lowest Common Multiple= 56

IgorLugansk [536]3 years ago

4 0

Answer:

56 seconds

Step-by-step explanation:

One way of doing this is to write out sequences for the number of seconds that elapse in each case.

First bulb: 0 7 14 28 56 63

Second bulb: 0 8 16 24 32 40 48 56

Both bulbs flicker after 56 seconds.

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If a = 4 units and b = 3 units, the length of the diagonal of the outside square rounded to the nearest tenth is

ANTONII [103]

Use pythagorean theorem
a^2+b^2=c^2
4^2+3^2=c^2
16+9=c^2
25=c^2
c=5
hope this helps!

7 0

3 years ago

Read 2 more answers

The first side of a triangle is 3 m shorter than the second side. The third side is 4 times as long as the first side. The perim

arsen [322]

I would have to say 2.3 but i dont know the real answer thats what i calculated

8 0

3 years ago

Write an equation in general form for the line passing through (-12,-1) and perpendicular to the line whose equation is 6x-y-4=0

Rus_ich [418]

The Solution:

Given the equation of a line:

$6x-y-4=0$

We are required to find the equation of a line that is perpendicular to the given line and passes through the point (-12,-1).

Step 1:

Determine the slope of 6x-y-4=0.

$\begin{gathered} \text{ Express y in terms of x.} \\ 6x-4=y \\ y=6x-4 \end{gathered}$

The coefficient x, after making y the subject of the formula, is the slope of the given line. That is:

$\begin{gathered} \text{ The slope of }6x-y-4=0\text{ is 6} \\ \\ Slope=m_1=6 \end{gathered}$

Step 2:

Find the slope of the required line:

Since the two lines are perpendicular, the slope is:

$\begin{gathered} m_2=\frac{-1}{m_1}=\frac{-1}{6} \\ \end{gathered}$

Step 3:

Find the equation of the line that passes through point (-12,-1)

By formula,

$y-y_1=m_2(x-x_1)$

In this case,

$\begin{gathered} x_1=-12 \\ y_1=-1 \end{gathered}$

Substituting these values in the formula, we get the required equation is:

$\begin{gathered} y--1=\frac{-1}{6}(x--12) \\ \\ y+1=-\frac{1}{6}(x+12) \end{gathered}$

Cross multiplying, we get

$\begin{gathered} 6y+6=-x-12 \\ \\ 6y+x+6+12=0 \\ \\ 6y+x+18=0 \end{gathered}$

Therefore, the correct answer is:

$6y+x+18=0$

5 0

1 year ago

Pre-image point N(6,-3) was dilated to point N'(2, -1). What was the scale factor used?

Aleksandr [31]

Answer:

scale factor = $\frac{1}{3}$

Step-by-step explanation:

Consider the ratio of the x and y- coordinates, image to original, that is

scale factor = $\frac{2}{6}$ = $\frac{-1}{-3}$ = $\frac{1}{3}$

8 0

3 years ago

What is the equation of the line?

Rudik [331]

Answer: It’s B) y=-1/3x-2

Step-by-step explanation:

5 0

3 years ago