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mrs_skeptik [129]

2 years ago

7

Can someone help me asap!!!

1 answer:

amm18122 years ago

5 0

Answer:

<em>Negative</em>

Step-by-step explanation:

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Simplify <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B5%20%2B%20%20%5Csqrt%7B7%7D%20%7D%7B3%20%2B%20%20%5Csqrt%7B7%7D%20%7

cricket20 [7]

Step-by-step explanation:

Times the eqn by 3-sqrt(7)/3-sqrt(7)

5 0

3 years ago

5. When the same side is part of two different triangles, you can use the _________ property to say that that side is congruent

dexar [7]

Answer:

I will give an answer if you answer mine

5 0

3 years ago

Graph triangle RST with vertices R(3, 7), S(-5, -2), and T(3, -5) and its image after a reflection over x = -3.

kirill115 [55]

Given:

The vertices of a triangle are R(3, 7), S(-5, -2), and T(3, -5).

To find:

The vertices of the triangle after a reflection over x = -3 and plot the triangle and its image on the graph.

Solution:

If a figure reflected across the line x=a, then

$(x,y)\to (-(x-a)+a,y)$

$(x,y)\to (-x+a+a,y)$

$(x,y)\to (2a-x,y)$

The triangle after a reflection over x = -3. So, the rule of reflection is

$(x,y)\to (2(-3)-x,y)$

$(x,y)\to (-6-x,y)$

The vertices of triangle after reflection are

$R(3,7)\to R'(-6-3,7)$

$R(3,7)\to R'(-9,7)$

Similarly,

$S(-5,-2)\to S'(-6-(-5),-2)$

$S(-5,-2)\to S'(-6+5,-2)$

$S(-5,-2)\to S'(-1,-2)$

And,

$T(3,-5)\to T'(-6-3,-5)$

$T(3,-5)\to T'(-9,-5)$

Therefore, the vertices of triangle after reflection over x=-3 are R'(-9,7), S'(-1,-2) and T'(-3,-5).

3 0

3 years ago

Finding Hypotenuse Lengths. Find the length of the hypotenuse.

ioda

Answer:

11) $\frac{3\sqrt{5}}{2}$

12) $4\sqrt{5}$

13) $22$

14) $18$

15) $2$

16) $24\sqrt{3}$

Step-by-step explanation:

For these problems I used the pythagorean theorem: $a^{2}+ b^{2}= c^{2}$ and SOHCAHTOA

Sin = $\frac{opposite}{hypotenuse}$

Cos = $\frac{adjacent}{hypotenuse}$

Tan = $\frac{opposite}{hypotenuse}$

11)

First find the length of the bottom side by using Cos

$cos(60)=\frac{x}{3}$

$3(cos(60))=x$

$1.5=x$

Then plug it into the formula for the pythagorean theorem to find the hypotenuse

$3^{2}+ 1.5^{2}= c^{2}$

$9+2.25=c^{2}$

$\sqrt{11.25}=\sqrt{c^{2}}$

$\frac{3\sqrt{5}}{2}$

12)

Find the length of the bottom side using Cos

$cos(60)=\frac{x}{8}$

$8(cos(60))=x$

$4=x$

Then plug it into the formula for the pythagorean theorem to find the hypotenuse

$8^{2}+ 4^{2}= c^{2}$

$64+16=c^{2}$

$\sqrt{80} =\sqrt{c^{2} }$

$4\sqrt{5}$

13)

Find the length of the other side by using Tan

$tan(30)=\frac{x}{11\sqrt{3} }$

$11\sqrt{3}* (tan(30)=x$

$11=x$

Then plug it into the formula for the pythagorean theorem to find the hypotenuse

$(11\sqrt{3}) ^{2}+ 11^{2}= c^{2}$

$363+121=c^{2}$

$\sqrt{484} =\sqrt{c^{2}}$

$22$

14)

(This is probably an easier way to do these problems)

Find the hypotenuse by using Cos ( $\frac{adjacent}{hypotenuse}$ )

$cos(60)=\frac{9}{x}$

$cos(60)x=9$

$x=\frac{9}{cos(60)}$

$x=18$

15)

Find the hypotenuse using Sin ( $\frac{opposite}{hypotenuse}$ )

$sin(30)=\frac{1}{x}$

$sin(30)x=1\\x=\frac{1}{sin(x)}$

$x=2$

16)

Find the hypotenuse using Cos ( $\frac{adjacent}{hypotenuse}$ )

$cos(60)=\frac{12\sqrt{3} }{x}$

$cos(60)x=12\sqrt{3}$

$x=\frac{12\sqrt{3} }{cos(60)}$

$x=24\sqrt{3}$

5 0

4 years ago

If 14 out of 20 like chocolate then how many people dislike chocolate?

Masteriza [31]

You know from the given information, there are 20 people in total.
When asked 14 people stated that they liked chocolate. This leaves a group of people out of 20 who do not like chocolate.

20-14=6

You know from this that 6 people do not like chocolate while 14 of them do

Hope I helped :)

6 0

3 years ago