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3 years ago

Please help dealing with missing angles

Mathematics

2 answers:

lbvjy [14]3 years ago

7 0

Uhm we’re the question also u can send a picture of it if u didn’t know

Goshia [24]3 years ago

5 0

Answer:

Where is the question.

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Please help it’s worth 50 points

blagie [28]

Answer:

17 hours 30 minutes

Step-by-step explanation:

3 0

4 years ago

Read 2 more answers

F. The relationship between x and y is not a straight line. <br> Does anyone know the answer??

Allisa [31]

Answer:

$y=\frac{1}{3}x-1$

As the line equation represents a straight line, so the relationship between x and y is a straight line.

Therefore, option E is true.

Step-by-step explanation:

Taking two points from the given line

(3, 0)

(-3, -2)

Finding the slope between two points

$\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(3,\:0\right),\:\left(x_2,\:y_2\right)=\left(-3,\:-2\right)$

$m=\frac{-2-0}{-3-3}$

$m=\frac{1}{3}$

From the graph, the y-intercept can be calculated by setting the x=0 and then check the corresponding value of y.

at x = 0, y=-1

Thus, the y-intercept = -1

We know that the slope-intercept form is

$y=mx+b$

where m is the slope, and b is the y-intercept

substituting the values of m=1/3 and the y-intercept b = -1

$y=mx+b$

$y=\frac{1}{3}x+\left(-1\right)$

$y=\frac{1}{3}x-1$

As the line equation represents a straight line, so the relationship between x and y is a straight line.

Therefore, option E is true.

5 0

3 years ago

Consider the graph of the linear function h(x) = –x + 5. Which could you change to move the graph down 3 units?

Feliz [49]

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

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the y-axis}$

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

with that template in mind, let's see

$\bf h(x)=-x+5\implies h(x)=\stackrel{A}{-1}(\stackrel{B}{1}x\stackrel{C}{+0})\stackrel{D}{+5}$

a shift down by 3 units, means a vertical shift downwards, so D needs to drop by 3 units.

$\bf h(x)=\stackrel{A}{-1}(\stackrel{B}{1}x\stackrel{C}{+0})\boxed{\stackrel{D}{+5-3}}\implies h(x)=-1(1x+0)+2
\\\\\\
h(x)=-x+2$

4 0

3 years ago

Please help!

Black_prince [1.1K]

It’s a bit blurry, but the first ones ratio is 6:4 yellow: red

7 0

3 years ago

Help please!!!!!!!!!!

adell [148]

Answer:

1/2?

Step-by-step explanation: