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3 years ago

Which points lie in Quadrant III? Choose all answers that are correct. A (–8, –2) B(–5, 4) C(–3, –9) D (4, –1) E (7, 0)

Mathematics

2 answers:

hodyreva [135]3 years ago

6 0

Answer:

The points A (-8, -2) and C(-3, -9) are in the quadrant III.

Step-by-step explanation:

The cartesian plane is divided into four sections named quadrants, the four quadrants are organized by an opposite direction to the hands of the clock (you can see it in the attachment).

Therefore, each quadrant has a characteristic on the sign of its numbers, so:

• The quadrant I has the characteristic that both numbers are positive.

• The quadrant II has the characteristic that the number that represents its location in X is negative and the number that represents it location in Y is positive.

• The quadrant III has the characteristic that both numbers are negative.

• The quadrant IV has the characteristic that the number that represents its location in X is positive and the number that represents it location in Y is negative.

Having in account the information, we can classified the points and be sure that the A and the C are in the quadrant III, because they are the only ones in which their both numbers are negative.

Studentka2010 [4]3 years ago

4 0

Quadrant 3 will have a negative x and a negative y........so ur answers are :
A and C

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In the pulley system shown in this figure, MQ = 30 mm, NP = 10 mm, and QP = 21 mm. Find MN.

ankoles [38]

<h2>Answer:</h2>

$\boxed{\overline{MN}=37.96}$

<h2>Step-by-step explanation:</h2>

For a better understanding of this problem, see the figure below. Our goal is to find $\overline{MN}$ . Since:

$\angle MRS=\angle MQP=90^{\circ} \\ \\ \overline{MQ}=\overline{MR}=30mm$

and $\overline{MN}$ is a common side both for ΔMRN and ΔMQN, then by SAS postulate, these two triangles are congruent and:

$\overline{RN}=\overline{QN}$

By Pythagorean theorem, for triangle NQP:

$\overline{QN}=\sqrt{\overline{NP}^2+\overline{QP}^2} \\ \\ \overline{QN}=\sqrt{10^2+21^2} \\ \\ \overline{QN}=\sqrt{541}$

Applying Pythagorean theorem again, but for triangle MQN:

$\overline{MN}=\sqrt{\overline{MQ}^2+\overline{QN}^2} \\ \\ \overline{MN}=\sqrt{30^2+(\sqrt{541})^2} \\ \\ \boxed{\overline{MN}=37.96}$

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3 years ago

Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

3 years ago

What’s the area for 16 and 8 3/4

Alborosie

Answer:

Area = 140

16 x 8.75 = 140

Step-by-step explanation:

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3 years ago