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3 years ago

C. Marcia made 10 bracelets for 5 friends. Jen made 12 bracelets

Mathematics

1 answer:

sveta [45]3 years ago

3 0

Answer:

Yes because before both of the numbers are equvilant and end the same

Step-by-step explanation:

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4. A driver set a goal to earn $2,000 in March. He receives a base salary of $1,000 per month as well as a 20% commission for al

VashaNatasha [74]

Answer:$1000

Step-by-step explanation:

3 0

3 years ago

70 POINTS!!!!!!!!!! Given: E, F, Q, D∈k(O),O ∈ ED, m∠DFQ = 10°, measure of arc EF = 28° Find: Angles of △EFQ

Likurg_2 [28]

Answer:

The measures of angles of triangle EFQ are

1) $m\angle EFQ=100\°$

2) $m\angle FEQ=66\°$

3) $m\angle EQF=14\°$

Step-by-step explanation:

step 1

Find the measure of arc QD

we know that

The inscribed angle is half that of the arc it comprises.

$m\angle DFQ=\frac{1}{2}(arc\ QD)$

substitute the given value

$10\°=\frac{1}{2}(arc\ QD)$

$20\°=(arc\ QD)$

$arc\ QD=20\°$

step 2

Find the measure of arc FQ

we know that

$arc\ QD+arc\ FQ+arc\ EF=180\°$ ---> because ED is a diameter (the diameter divide the circle into two equal parts)

substitute the given values

$20\°+arc\ FQ+28\°=180\°$

$arc\ FQ=180\°-48\°=132\°$

step 3

Find the measure of angle EFQ

we know that

The inscribed angle is half that of the arc it comprises.

$m\angle EFQ=\frac{1}{2}(arc\ QD+arc\ ED)$

substitute the given value

$m\angle EFQ=\frac{1}{2}(20\°+180\°)=100\°$

step 4

Find the measure of angle FEQ

we know that

The inscribed angle is half that of the arc it comprises.

$m\angle FEQ=\frac{1}{2}(arc\ FQ)$

substitute the given value

$m\angle FEQ=\frac{1}{2}(132\°)=66\°$

step 5

Find the measure of angle EQF

we know that

The inscribed angle is half that of the arc it comprises.

$m\angle EQF=\frac{1}{2}(arc\ EF)$

substitute the given value

$m\angle EQF=\frac{1}{2}(28\°)=14\°$

5 0

4 years ago

There are 154 players on 11 teams . How many players are on one team

jarptica [38.1K]

Answer:

there are 14 on a single team

Step-by-step explanation:

154÷11=14

Or you can do this

11 • 14

add 11 14 times you get 154

7 0

4 years ago

Solve: (3x^2-y)dx + (4y^3-x)dy =0 and find the solution passing through (1,1).

nordsb [41]

Step-by-step explanation:

The given equation is

$(3x^{2}-y)dx+(4y^{3}-x)dy=0\\M(x,y)dx+N(x,y)dy=0$

As a check for exactness we have

$\frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial (4y^{3}-x)}{\partial x} =-1\\\\\frac{\partial M}{\partial y}=\frac{\partial (3x^{3}-y)}{\partial y} =-1\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}=-1$

Hence the given equation is an exact differential equation and thus the solution is given by

thus the solution is given by

$u(x,y)=\int M(x,y)\partial x+\phi (y)\\\\u(x,y)=\int (3x^{2}-y)\partial x+\phi (y,c)\\\\u(x,y)=x^{3}-xy+\phi (y,c)\\\\$

Similarly we have

$u(x,y)=\int N(x,y)\partial y+\phi (x,c)\\\\u(x,y)=\int (4y^{3}-x)\partial y+\phi (x,c)\\\\u(x,y)=y^{4}-xy+\phi (x,c)\\\\$

Comparing both the solutions we infer

$\phi (x,c)=x^{3}+c$

Hence the solution becomes

$u(x,y)=x^{3}+y^{4}-xy=c$

given boundary condition is that it passes through (1,1) hence

$1^{3}+1^{4}-1=c\\\\\therefore c=1$

thus solution is

$u(x,y)=x^{3}+y^{4}-xy=1$

4 0

3 years ago

is there anyone that can solve this for me and tell me how they came up with the answer?? (pic included)

worty [1.4K]

So to begin, substitute all of the variables in for their corresponding value.

X=3
Y=-5
Z=-1

2XY+6YZ^2
2(3)(-5)+6(-5)(-1^2)
-30+-30
-60

Hopefully this is what they're asking for!

4 0

3 years ago