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3 years ago

Alguien que me ayude con esto

Mathematics

1 answer:

e-lub [12.9K]3 years ago

5 0

7, 42,28,30,8,36 and I don’t know the rest

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Solve the system by the elimination method. Check your work. R - 9S = 2 3R - 3S = -10

stiks02 [169]

-3(r-9s=2)+(3r-3s=-10)

24s=-16

s=-2/3, now using either original equation we can solve for r.

3r-3s=-10 becomes:

3r+2=-10

3r=-12

r=-4 so the solution is the point:

(s,r)=(-2/3, -4)

3 0

3 years ago

You focus your camera on a circular fountain. Your camera is at the vertex of the angel formed by tangents to the fountain. You

drek231 [11]

Answer:

The measure of the arc of the circular basin = 136°

Step-by-step explanation:

The measure of an angle formed when two line intercepts outside a circle is half the difference of the measure of the intercepted arcs.

Mathematically, the is represented as:

Measure of an angle = 1/2(big angle - Small angle)

This values are given in the question

Measure of an angle = Measure of angle formed by tangents to the fountain = 44°

big angle is represented by = 360°-x

small angle is represented by = x

Therefore, we have

44° = 1/2( 360° - x -x)

44° = 1/2(360° - 2x)

Cross multiply

44° × 2 = 360° - 2x

88° = 360° - 2x

88° - 360° = - 2x

-272° = -2x

x = -272/-2

x = 136°

The measure of the arc of the circular basin = 136°

6 0

3 years ago

The price of oranges went from $.50 per Ib to $1.80 per lb in four years. Find

Marizza181 [45]

Answer:

Step-by-step explanation:

8 0

3 years ago

Find an explicit solution of the given initial-value problem.dy/dx = ye^x^2, y(3) = 1.

tamaranim1 [39]

Answer:

$y=exp(\int\limits^x_4 {e^{-t^{2} } } \, dt)$

Step-by-step explanation:

This is a separable equation with an initial value i.e. y(3)=1.

Take y from right hand side and divide to left hand side ;Take dx from left hand side and multiply to right hand side:

$\frac{dy}{y} =e^{-x^{2} }dx$

Take t as a dummy variable, integrate both sides with respect to "t" and substituting x=t (e.g. dx=dt):

$\int\limits^x_3 {\frac{1}{y} } \, \frac{dy}{dt} dt=\int\limits^x_3 {e^{-t^{2} } } dt$

Integrate on both sides:

$ln(y(t))\left \{ {{t=x} \atop {t=3}} \right. =\int\limits^x_3 {e^{-t^{2} } } \, dt$

Use initial condition i.e. y(3) = 1:

$ln(y(x))-(ln1)=\int\limits^x_3 {e^{-t^{2} } } \, dt\\ln(y(x))=\int\limits^x_3 {e^{-t^{2} } } \, dt\\$

Taking exponents on both sides to remove "ln":

$y=exp (\int\limits^x_3 {e^{-t^{2} } } \, dt)$

7 0

3 years ago

Help with geometry question

IrinaVladis [17]

Answer:

figure b

Step-by-step explanation:

3 0

3 years ago