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3 years ago

15

A student thinks that (1, 2) is a solution to y = 2x - 3. While checking the solution, the student gets 1 = 2(2) -3. What error

did the student make when entering the values of x and y?

1 answer:

BabaBlast [244]3 years ago

6 0

He put in the value of x into y
and the value of y in x
if that confuse you here this is what it will look like
(x,y)->(y,x)

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2 years ago

I need to know the improper fractions answers for:

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Answer:

Part 1) $x=\frac{15}{2}\ units$

Part 2) $z=\frac{15\sqrt{3}}{2}\ units$

Part 3) $y= \frac{15\sqrt{3}}{4}\ units$

Part 4) $b= \frac{45}{4}\ units$

Step-by-step explanation:

step 1

Find the value of x

In the large right triangle

$cos(60^o)=\frac{x}{15}$ ----> by CAH (adjacent side divided by the hypotenuse)

Remember that

$cos(60^o)=\frac{1}{2}$

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$\frac{1}{2}=\frac{x}{15}$

solve for x

$x=\frac{15}{2}\ units$ ---> improper fraction

step 2

Find the value of z

In the large right triangle

Applying the Pythagorean Theorem

$15^2=x^2+z^2$

substitute the value of x

$15^2=(\frac{15}{2})^2+z^2$

solve for z

$z^2=15^2-(\frac{15}{2})^2$

$z^2=225-\frac{225}{4}$

$z^2=\frac{675}{4}$

$z=\frac{\sqrt{675}}{2}\ units$

simplify

$z=\frac{15\sqrt{3}}{2}\ units$

step 3

Find the value of y

In the right triangle of the right

$sin(30^o)=\frac{y}{z}$ ---> by SOH (opposite side divided by the hypotenuse)

substitute the given values of y and z

Remember that

$sin(30^o)=\frac{1}{2}$

so

$\frac{1}{2}=y:\frac{15\sqrt{3}}{2}$

solve for y

$\frac{1}{2}= \frac{2y}{15\sqrt{3}}$

$y= \frac{15\sqrt{3}}{4}\ units$

step 4

Find the value of b

In the right triangle of the right

$cos(30^o)=\frac{b}{z}$ ---> by CAH (adjacent side divided by the hypotenuse)

substitute the given values of y and z

Remember that

$cos(30^o)=\frac{\sqrt{3}}{2}$

so

$\frac{\sqrt{3}}{2}=b:\frac{15\sqrt{3}}{2}$

solve for y

$\frac{\sqrt{3}}{2}= \frac{2b}{15\sqrt{3}}$

$b= \frac{45}{4}\ units$

7 0

2 years ago