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3 years ago

Using Trig to find angles !! help please

Mathematics

1 answer:

notsponge [240]3 years ago

7 0

Answer:

56.1°

Step-by-step explanation:

$\tan \:x \degree = \frac{64}{43} \\ \tan \:x \degree = 1.48837209 \\ x \degree = { \tan}^{ - 1} (1.48837209) \\ x \degree = 56.103833383 \degree \\ x =56.1 \degree$

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State whether set A and B are equal, equivalent, both, or neither. A = {9, 8, 7} B = {8, 9, 10}

bearhunter [10]

They're just equivalent.

7 0

2 years ago

Priya filled 5 jars, using a total of 7 ½ cups of strawberry jam. How many cups of jam are in each jar?

Leya [2.2K]

Answer:

1 1/2 cups of jam per jar

Step-by-step explanation:

Divide 7 1/2 by 5 to get 1 1/2 cups of jam per jar.

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2 years ago

If we measured the heights of different plants and got 50”, 55”, 48”, 50” and 52”. What

mr Goodwill [35]

Answer:

Step-by-step explanation:

50+55+48+50+52=255

255/5=51

Anwser 51

5 0

2 years ago

If an hour is 45 years how long is a minute

Ainat [17]

Since one hour = 45 years, one minute should be 1/60th of 45 years because 60 minutes = an hour.

45/60 = 3/4

So one minute would be 3/4 of an hour, or 45 minutes.

Hope this helps!

8 0

2 years ago

Match the circle equations in general form with their corresponding equations in standard form. Not all will be used.

Xelga [282]

The standard form of the equation of a circumference is given by the following expression:

 $(x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius$

On the other hand, the general form is given as follows:

 $x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}$ 

In this way, we can order the mentioned equations as follows:

Equations in Standard Form:

 $\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46$

Equations in General Form:

$\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0$

$\bold{6)} \ x^{2}+y^{2}+2x-12y$

So let's match each equation:

$\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4$

Then, its general form is:

$x^{2}+y^{2}-12x-8y-4=0$

First. a) matches 5) 

$\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20$

Then, its general form is:

$x^{2}+y^{2}-4x+12y-20=0$

Second. b) matches 1) 

$\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5$

Then, its general form is:

$x^{2}+y^{2}+4x+6y-5=0$

Third. c) matches 3)

$\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9$

Then, its general form is: $x^{2}+y^{2}+2x-12y-9=0$

Fourth. d) matches 6)

6 0

3 years ago

Read 2 more answers