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2 years ago

Help y’all!! it’s gradpoint!! this is the last question & i’ve failed this test so many times :(

Mathematics

1 answer:

scoray [572]2 years ago

5 0

Answer:

just study study study

Step-by-step explanation:

you will learned

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In the adjoining figure, find the value of x for which<br>The lines l and m are parallel.

Andrew [12]

hope it helps I tried my best

8 0

4 years ago

Label the trapezoids with the given side measures: , and Use similar figures and the given side lengths to complete the followin

Anna71 [15]

Answer:A trapezoid has ONLY ONE set of parallel sides. When proving a figure is a trapezoid, it is necessary to prove ... The angles on the same side of a leg are called adjacent angles such as ∠A and ∠D are ... Image: The midsegment of a trapezoid is: parallel to both bases has length ... Given : ABCD is an isosceles trapezoid.

Step-by-step explanation:

4 0

3 years ago

Can someone help right away!! please!!

larisa [96]

Answer:

y = - 2x + 1

Step-by-step explanation:

Substitute an x value from the table into the equation

x = 1 : y = - 2(1) = - 2 ← require to add 1 to obtain y = - 1

x = 4 : y = - 2(4) = - 8 ← require to add 1 yo obtain y = - 7

Then equation describing the relationship is

y = - 2x + 1

4 0

3 years ago

Please Help ASAP

Orlov [11]

The C(x) equation is a parabola with a squared "x" and a positive leading term coefficient, meaning, is a vertical parabola and it opens upwards

the lowest point, or lowest cost value, it reaches, is the U-turn point, or vertex

$\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{llccll}
y = &{{ 1}}x^2&{{ -320}}x&{{ +40052}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

so, the lowest cost is at $\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}$

8 0

3 years ago

Graph the following sequence of transformations for

Sedbober [7]

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

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

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

with that template in mind.

$\bf f(x)=\sqrt[3]{x}\qquad \boxed{-f(x+2)+4}\implies f(x)=\stackrel{A}{-1}\sqrt[3]{\stackrel{B}{1}x\stackrel{C}{+2}}\stackrel{D}{+4}\qquad 
\\\\\\
f(x)=-\sqrt[3]{x+2}+4$

A = -1, reflected over the x-axis

B = 1, C = 2, horizontal shift of C/B or 2, to the left.

D = 4, shifted upwards by 4 units.

check the picture below.

7 0

3 years ago