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3 years ago

Help ill give brainlest pls

Mathematics

2 answers:

QveST [7]3 years ago

6 0

Answer= D
x values don’t change

Dafna1 [17]3 years ago

3 0

Answer:

The answer is D

Step-by-step explanation:

This is because the x value didn't change while the y values became negative.

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PLEASEEEE HELPPP ME this is due today!!!

Sati [7]

The step after the given would be subtraction

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3 years ago

Read 2 more answers

Help with this question!

zheka24 [161]

Since AS is a height issued from A and the perpendicular bisector of [MP] at the same time (given), so the triangle AMP is an isosceles triangle of vertex A. Then, AM=AP
MS=SP ( AS bisects MP as stated in the given )
AS is a common side between triangles ASM and ASP
Therefore, triangles ASM and ASP are congruent (SSS)

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4 years ago

The ratio of strawberries to bananas in a smoothie recipe are shown in the graph below.

dedylja [7]

Answer: The last table is the correct answer.

Step-by-step explanation: The ratio is 6 strawberries to 1 banana.

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3 years ago

Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane

liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

Answer:

$Rate = 0.935042^\circ /cm$

Step-by-step explanation:

Given

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$T(x,y) =x\sin2y$

$r = 1m$

$v = 2m/s$

Express the given point P as a unit tangent vector:

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$u = \frac{\sqrt 3}{2}i - \frac{1}{2}j$

Next, find the gradient of P and T using: $\triangle T = \nabla T * u$

Where

$\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})} = (sin \sqrt 3)i + (cos \sqrt 3)j$

So: the gradient becomes:

$\triangle T = \nabla T * u$

$\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] * [\frac{\sqrt 3}{2}i - \frac{1}{2}j]$

By vector multiplication, we have:

$\triangle T = (sin \sqrt 3)* \frac{\sqrt 3}{2} - (cos \sqrt 3) \frac{1}{2}$

$\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)$

$\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5$

$\triangle T = 0.935042$

Hence, the rate is:

$Rate = \triangle T = 0.935042^\circ /cm$

3 0

3 years ago

The world population at the beginning of 1980 was 4.5 billion. Assuming that the population continued to grow at the rate of app

AfilCa [17]

Answer:

$Q(t) = 4.5(1.013)^{t}$

The world population at the beginning of 2019 will be of 7.45 billion people.

Step-by-step explanation:

The world population can be modeled by the following equation.

$Q(t) = Q(0)(1+r)^{t}$

In which Q(t) is the population in t years after 1980, in billions, Q(0) is the initial population and r is the growth rate.

The world population at the beginning of 1980 was 4.5 billion. Assuming that the population continued to grow at the rate of approximately 1.3%/year.

This means that $Q(0) = 4.5, r = 0.013$

$Q(t) = Q(0)(1+r)^{t}$

$Q(t) = 4.5(1.013)^{t}$

What will the world population be at the beginning of 2019 ?

2019 - 1980 = 39. So this is Q(39).

$Q(t) = 4.5(1.013)^{t}$

$Q(39) = 4.5(1.013)^{39} = 7.45$

The world population at the beginning of 2019 will be of 7.45 billion people.

6 0

3 years ago