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2 years ago

9

How many angles must be supplementary with angle 14? 4 lines intersect to form 16 angles. The angles created, clockwise from top

left are 1, 2, 3, 4; 5, 6, 7, 8; 13, 14, 15, 16; 9, 10, 11, 12.
one

two

four

eight

2 answers:

nasty-shy [4]2 years ago

8 0

It’s 4.................... ..........................

trapecia [35]2 years ago

4 0

Answer:

four

Step-by-step expif this is worng but i had a similar question and that as the answer

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Solve the system of equations algebraically.

Romashka [77]

A should be the right answer because 4*5/4=5 + 5*2=10 and 5+10=15.

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2 years ago

after a birthday party, 1 4 of a large cake was left. If this piece of cake was shared equally among the 6 people, what fraction

il63 [147K]

The 6 people each got $\frac{1}{24}$ . To get this answer you need to divide the $\frac{1}{4}$ of the cake among 6 people.

Equation ⇒ $\frac{1}{4}$ x $\frac{1}{6}$

You get the answer by multiplying the bottom denominators.

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2 years ago

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If you say that you can help me with homework then why can't I make a code to join and I had to pick one to send this sincerely,

Nataly_w [17]

Answer:

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Step-by-step explanation:

6 0

3 years ago

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Tamara finds the sum of two number cubes rolled at the same time. The chart below shows all possible sums from the 36 possible c

Georgia [21]

Answer:

18 times

Step-by-step explanation:

Sample space ; roll of 2 number cubes = 6^2 = 36

Sum of 4 in 2 cube rolls = 3

Probability = required outcome / Total possible outcomes

P(sum of 4) = 3 / 36 = 1 / 12

If number cube is rolled 216 times :

Probability on 1 roll * Number of rolls

(1 / 12) * 216

0.0833333 * 216

= 18

3 0

2 years ago

Please I need help with differential equation. Thank you

Inga [223]

1. I suppose the ODE is supposed to be

$\mathrm dt\dfrac{y+y^{1/2}}{1-t}=\mathrm dy(t+1)$

Solving for $\dfrac{\mathrm dy}{\mathrm dt}$ gives

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{y+y^{1/2}}{1-t^2}$

which is undefined when $t=\pm1$ . The interval of validity depends on what your initial value is. In this case, it's $t=-\dfrac12$ , so the largest interval on which a solution can exist is $-1\le t\le1$ .

2. Separating the variables gives

$\dfrac{\mathrm dy}{y+y^{1/2}}=\dfrac{\mathrm dt}{1-t^2}$

Integrate both sides. On the left, we have

$\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\int\frac{\mathrm dz}{z+1}$

where we substituted $z=y^{1/2}$ - or $z^2=y$ - and $2z\,\mathrm dz=\mathrm dy$ - or $\mathrm dz=\dfrac{\mathrm dy}{2y^{1/2}}$ .

$\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\ln|z+1|=2\ln(y^{1/2}+1)$

On the right, we have

$\dfrac1{1-t^2}=\dfrac12\left(\dfrac1{1-t}+\dfrac1{1+t}\right)$

$\displaystyle\int\frac{\mathrm dt}{1-t^2}=\dfrac12(\ln|1-t|+\ln|1+t|)+C=\ln(1-t^2)^{1/2}+C$

So

$2\ln(y^{1/2}+1)=\ln(1-t^2)^{1/2}+C$

$\ln(y^{1/2}+1)=\dfrac12\ln(1-t^2)^{1/2}+C$

$y^{1/2}+1=e^{\ln(1-t^2)^{1/4}+C}$

$y^{1/2}=C(1-t^2)^{1/4}-1$

I'll leave the solution in this form for now to make solving for $C$ easier. Given that $y\left(-\dfrac12\right)=1$ , we get

$1^{1/2}=C\left(1-\left(-\dfrac12\right)^2\right))^{1/4}-1$

$2=C\left(\dfrac54\right)^{1/4}$

$C=2\left(\dfrac45\right)^{1/4}$

and so our solution is

$\boxed{y(t)=\left(2\left(\dfrac45-\dfrac45t^2\right)^{1/4}-1\right)^2}$

3 0

3 years ago