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2 years ago

Segments BG and AH are lines are what. A.skew B. Perpendicular C. Parallel

Mathematics

1 answer:

solmaris [256]2 years ago

7 0

Answer:

A, they are skew

Step-by-step explanation:

Skew means that they are not perpendicular and they are not parallel.

Sorry if it turns out wrong.

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If a is an integer and n is a positive integer, we define a mod n to be the remainder when a is divided by n. The integer n is c

dezoksy [38]

Answer:

Divisor

Step-by-step explanation:

Given

$a\ mod\ n$

Required

What does n represents

In arithmetic, the number that divides another is referred to as divisor;

Take for instance;

$5\ mod\ 3 = 2$

This implies that when 5 is divided by 3, the remainder is 2

The bold text (divided by 3) means that 3 is the divisor of 5;

When compared to $a\ mod\ n$

We have that

n = 3

<em>Reason being that, n and 3 are in the same position base on order of appearance;</em>

Hence, n represents the divisor

4 0

2 years ago

you are creating a model of the solar system and an asteroid that has a linear path that travels through it. The sun is at the c

Anettt [7]

Answer:

still thinking

Step-by-step explanation:

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2 years ago

Compute the amount of interest earned in the following simple interest problem. A deposit of $1,600 at 6% for 180 days =

Nikitich [7]

6% of $1,600 is = 96 if you multiple it with 180 : 96 × 180 = 17,280

7 0

3 years ago

Read 2 more answers

Solution to m^2- 36=0

alex41 [277]

Answer:

m=(-6,6)

Step-by-step explanation:

m^2 -36 = 0

Reorder the terms:

-36 + m^2 = 0

Solving for variable 'm'.

Add '36' to each side of the equation.

-36 + 36 + m^2 = 0 + 36

Combine like terms: -36 + 36 = 0

0 + m^2 = 0 + 36

m^2 = 0 + 36

Combine like terms: 0 + 36 = 36

m^2 = 36

Simplifying

m^2 = 36

Take the square root of each side:

√m^2=+/-√36

m=(+/-)6

m = {-6, 6}

4 0

2 years ago

Write an equation that has a graph with the shape of y = x^2, but shifted left 3 units.

svetoff [14.1K]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf % template detailing
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\$

$\bf \bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

now, with that template in mind, let's see

$\bf y=x^2\implies 
\begin{array}{llll}
y=(&1x&-0)^2\\
&B&C
\end{array}$

so, just change C to +3, thus C/B is 3/1

6 0

2 years ago