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Bess [88]
3 years ago
7

1. y varies directly with x, and y = 7 when x = 2. What is the value of y when x = 5?

Mathematics
1 answer:
faltersainse [42]3 years ago
4 0
1. y=mx where m is the constant

7/2 = y/5
2y = 35
y = 35/2
y = 17.5


2. y=mx where m is the constant

5=m4
m= 5/4
8=5/4 x
x= 6.4
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Lines m and pare parallel. If the slope of line mis
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Answer:

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3 0
3 years ago
Line a is parallel to line b, m∠1=2x+44 and m∠ 5=5x+38 Find the value of x.
Mumz [18]
<span>Line a is parallel to line b:
</span><span>m∠1=m∠ 5, because they are correspondent angles, then:
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3 years ago
Is this enough information for Angelina to find the interest rate? Explain.
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6 0
4 years ago
I dont understand this question
Stells [14]

A parachutist descends 65 feet in 5 seconds.

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4 0
3 years ago
Read 2 more answers
Help me pls! Thank you so much
frosja888 [35]
\bf cos\left[tan^{-1}\left(\frac{12}{5}  \right)+ tan^{-1}\left(\frac{-8}{15}  \right) \right]\\&#10;\left. \qquad  \qquad  \quad   \right.\uparrow \qquad \qquad  \qquad  \uparrow \\&#10;\left. \qquad  \qquad  \quad   \right.\alpha \qquad \qquad  \qquad  \beta&#10;\\\\\\&#10;\textit{that simply means }tan(\alpha)=\cfrac{12}{5}\qquad and\qquad tan(\beta)=\cfrac{-8}{5}&#10;\\\\\\&#10;\textit{so, we're really looking for }cos(\alpha+\beta)

now.. hmmm -8/15  is rather ambiguous, since the negative sign is in front of the rational, and either 8 or 15 can be negative, now, we happen to choose the 8 to get the minus, but it could have been 8/-15

ok, well hmm so, the issue boils down to 

\bf tan(\theta)=\cfrac{opposite}{adjacent}\qquad thus&#10;\\\\\\&#10;tan(\alpha)=\cfrac{12}{5}\cfrac{\leftarrow opposite=b}{\leftarrow  adjacent=a}&#10;\\\\\\&#10;\textit{so, what is the hypotenuse "c"?}\\&#10;\textit{ well, let's use the pythagorean theorem}&#10;\\\\\\&#10;c=\sqrt{a^2+b^2}\implies c=\sqrt{25+144}\implies c=\sqrt{169}\implies \boxed{c=13}\\\\&#10;-----------------------------\\\\&#10;\textit{this simply means }\boxed{cos(\alpha)=\cfrac{5}{13}\qquad \qquad sin(\alpha)=\cfrac{12}{13}&#10;}


now, let's take a peek at the second angle, angle β

\bf tan(\beta)=\cfrac{-8}{15}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a}&#10;\\\\\\&#10;\textit{again, let's find "c", or the hypotenuse}&#10;\\\\\\&#10;c=\sqrt{15^2+(-8)^2}\implies c=\sqrt{289}\implies \boxed{c=17}\\\\&#10;-----------------------------\\\\&#10;thus\qquad \boxed{cos(\beta)=\cfrac{15}{17}\qquad \qquad sin(\beta)=\cfrac{-8}{17}}

now, with that in mind, let's use the angle sum identity for cosine

\bf cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}})\\\\&#10;-----------------------------\\\\&#10;cos({{ \alpha}} + {{ \beta}})= \left( \cfrac{5}{13} \right)\left( \cfrac{15}{17} \right)-\left( \cfrac{12}{13} \right)\left( \cfrac{-8}{17} \right)&#10;\\\\\\&#10;cos({{ \alpha}} + {{ \beta}})= \cfrac{75}{221}-\cfrac{-96}{221}\implies cos({{ \alpha}} + {{ \beta}})= \cfrac{75}{221}+\cfrac{96}{221}&#10;\\\\\\&#10;\boxed{cos({{ \alpha}} + {{ \beta}})=\cfrac{171}{221}}

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3 years ago
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