All categories

3 years ago

The measures of two complementary angles are 6y+3 and 4y−13 . Find the measures of the angles.

Mathematics

2 answers:

Cerrena [4.2K]3 years ago

8 0

Y = 10 degrees

work:
6y + 3 + 4y - 13 = 90
6y + 4y = 90 + 13 - 3
10y = 100
y = 10

photoshop1234 [79]3 years ago

4 0

63° and 27°

2 complementary angles sum to 90°, hence

6y + 3 + 4y - 13 = 90

10y - 10 = 90 ( add 10 to both sides )

10y = 100 ( divide both sides by 10 )

y = 10

The angles are

6y + 3 = (6 × 10 ) + 3 = 60 + 3 = 63°

4y - 13 = (4 × 10 ) - 13 = 40 - 13= 27°

note that 63° + 27° = 90°

You might be interested in

How many distinguishable ways can you arrange the letters in the word "MISSISSIPPI"?

Ipatiy [6.2K]

34,650 is the answer, I think

3 0

3 years ago

Read 2 more answers

How do you solve for what is a? Please answer quickly?<br><br> 4(a-2)=3(a+4)

Lunna [17]

4(a-2)=3(a+4)
4a-8=3a+12
a=20

8 0

3 years ago

How would you estimate it?

likoan [24]

I would just do for the first one 7/16 - 3/16 = 4/16 which is also 1/4 I'm not sure ho2 you estimate it

7 0

3 years ago

Someone help me with this ASAP!!!

kramer

the new price of the bike is $380

7 0

3 years ago

The radius of a spherical balloon is measured as 20 inches, with a possible error of 0.03 inch. Use differentials to approximate

soldi70 [24.7K]

Answer:

a) $V = 33510.322\,in^{3}$ , b) $A_{s} = 5026.548\,in^{2}$ , c) $\% V = 0.450\,\%$ , $\%A_{s} = 0.300\,\%$ .

Step-by-step explanation:

The volume and the surface area of the sphere are, respectively:

$V = \frac{4}{3}\pi \cdot r^{3}$

$A_{s} = 4\pi \cdot r^{2}$

a) The volume of the sphere is:

$V = \frac{4}{3}\pi \cdot (20\,in)^{3}$

$V = 33510.322\,in^{3}$

b) The surface area of the sphere is:

$A_{s} = 4\pi \cdot (20\,in)^{2}$

$A_{s} = 5026.548\,in^{2}$

c) The total differentials for volume and surface area of the sphere are, respectively:

$\Delta V = 4\pi\cdot r^{2}\,\Delta r$

$\Delta V = 4\pi \cdot (20\,in)^{2}\cdot (0.03\,in)$

$\Delta V = 150.796\,in^{3}$

$\Delta A_{s} = 8\pi\cdot r \,\Delta r$

$\Delta A_{s} = 8\pi \cdot (20\,in)\cdot (0.03\,in)$

$\Delta A_{s} = 15.080\,in^{2}$

Relative errors are presented hereafter:

$\%V = \frac{\Delta V}{V}\times 100\%$

$\%V = \frac{150.796 \,in^{3}}{33510.322\,in^{3}}\times 100\,\%$

$\% V = 0.450\,\%$

$\% A_{s} = \frac{\Delta A_{s}}{A_{s}}\times 100\,\%$

$\% A_{s} = \frac{15.080\,in^{2}}{5026.548\,in^{2}}\times 100\,\%$

$\%A_{s} = 0.300\,\%$

4 0

3 years ago