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2 years ago

Find the measure of the missing angles.

Mathematics

1 answer:

Rainbow [258]2 years ago

7 0

Step-by-step explanation:

c=144°(vertically opposite angle)

b+c=180°(straight angle)

b+144=180

b=180-144

b=36°

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In this exercise, find the area of the shaded part.

ivann1987 [24]

Since that angle is 45 degrees, the white triangle is an isosceles triangle. Therefore, the hight of the big triangle is equal to 12. The are of the shaded triangle is 4x12/2 = 24

5 0

2 years ago

13. On a trip of 525 miles, how many minutes faster is it to travel at 60 miles per hour than

lara [203]

I have no clue hahaha

5 0

3 years ago

Mr. Wilson bought a bag of birdseed and put half of it in his bird feeder. He split the other half equally among his 4 pet birds

pickupchik [31]

Answer:

each pet bird got 1/8 of the bag of birdseed

Step-by-step explanation:

the easiest way to do this is to divide the bag into eighths. if you divide the bag into eighths, 4/8 of the bag, being half, goes into the bird feeder. the other 4/8 of the bag is divided among Mr. Wilson's four pet birds, leaving each bird to get 1/4 of the bag when split evenly.

hope this helps

5 0

2 years ago

Find the sum of the first 25 terms in this geometric series: 8 + 6 + 4.5...

Ksivusya [100]

Step-by-step explanation:

Given the geometric sequence

8 + 6 + 4.5...

A geometric sequence has a constant ratio and is defined by

$a_n=a_1\cdot r^{n-1}$

$\mathrm{Compute\:the\:ratios\:of\:all\:the\:adjacent\:terms}:\quad \:r=\frac{a_{n+1}}{a_n}$

$\frac{6}{8}=\frac{3}{4},\:\quad \frac{4.5}{6}=\frac{3}{4}$

$\mathrm{The\:ratio\:of\:all\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}$

$r=\frac{3}{4}$

$\mathrm{The\:first\:element\:of\:the\:sequence\:is}$

$a_1=8$

$\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}\:$

$a_n=8\left(\frac{3}{4}\right)^{n-1}$

$\mathrm{Geometric\:sequence\:sum\:formula:}$

$a_1\frac{1-r^n}{1-r}$

$\mathrm{Plug\:in\:the\:values:}$

$n=25,\:\spacea_1=8,\:\spacer=\frac{3}{4}$

$=8\cdot \frac{1-\left(\frac{3}{4}\right)^{25}}{1-\frac{3}{4}}$

$\mathrm{Multiply\:fractions}:\quad \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}$

$=\frac{\left(1-\left(\frac{3}{4}\right)^{25}\right)\cdot \:8}{1-\frac{3}{4}}$

$=\frac{8\left(-\left(\frac{3}{4}\right)^{25}+1\right)}{\frac{1}{4}}$

$\mathrm{Apply\:exponent\:rule}:\quad \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}$

$=\frac{8\left(-\frac{3^{25}}{4^{25}}+1\right)}{\frac{1}{4}}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{\frac{b}{c}}=\frac{a\cdot \:c}{b}$

$=\frac{\left(1-\frac{3^{25}}{4^{25}}\right)\cdot \:8\cdot \:4}{1}$

$\mathrm{Multiply\:the\:numbers:}\:8\cdot \:4=32$

$=\frac{32\left(-\frac{3^{25}}{4^{25}}+1\right)}{1}$

$=\frac{32\cdot \frac{4^{25}-3^{25}}{4^{25}}}{1}$ ∵ $\mathrm{Join}\:1-\frac{3^{25}}{4^{25}}:\quad \frac{4^{25}-3^{25}}{4^{25}}$

$=32\cdot \frac{4^{25}-3^{25}}{4^{25}}$

$=\frac{\left(4^{25}-3^{25}\right)\cdot \:32}{4^{25}}$

$=\frac{2^5\left(4^{25}-3^{25}\right)}{2^{50}}$ ∵ $\mathrm{Factor}\:32:\ 2^5$ , $\mathrm{Factor}\:4^{25}:\ 2^{50}$

$=\frac{4^{25}-3^{25}}{2^{45}}$ ∵ $\mathrm{Cancel\:}\frac{\left(4^{25}-3^{25}\right)\cdot \:2^5}{2^{50}}:\quad \frac{4^{25}-3^{25}}{2^{45}}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a\pm \:b}{c}=\frac{a}{c}\pm \frac{b}{c}$

$=\frac{4^{25}}{2^{45}}-\frac{3^{25}}{2^{45}}$

$=32-\frac{3^{25}}{2^{45}}$ ∵ $\frac{4^{25}}{2^{45}}=32$

$=32-0.024$ ∵ $\frac{3^{25}}{2^{45}}=0.024$

$=31.98$

Therefore, the sum of the first 25 terms in this geometric series: 31.98

3 0

3 years ago

Two particles are fixed to an x axis: particle 1 of charge -1.00 x 10-7 C is at the origin and particle 2 of charge +1.00 x 10-7

r-ruslan [8.4K]

The electric field strength at any point from a charged particle is given by E = kq/r^2 and we can use this to calculate the field strength of the two fields individually at the midpoint.

The field strength at midway (r = 0.171/2 = 0.0885 m) for particle 1 is E = (8.99x10^9)(-1* 10^-7)/(0.0885)^2 = -7.041 N/C and the field strength at midway for particle 2 is E = (8.99x10^9)(5.98* 10^-7)/(0.0935)^2 = -7.041 N/C

Note the sign of the field for particle 1 is negative so this is attractive for a test charge whereas for particle 2 it is positive therefore their equal magnitudes will add to give the magnitude of the net field, 2*7.041 N/C = 14.082 N/C

6 0

3 years ago