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3 years ago

144 is the product of -12 and q

Mathematics

2 answers:

Luba_88 [7]3 years ago

7 0

A is equal to -12 as two negatives multiplied together form a positive and 12 times 12 is 144

koban [17]3 years ago

4 0

Answer:

Step-by-step explanation:

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Fill in the missing information. Tim Worker is doing his budget. He discovers that the average miscellaneous expense is $45.00 w

nlexa [21]

Answer:

$P(38.6$

And we can assume a normal distribution and then we can solve the problem with the z score formula given by:

$z=\frac{X -\mu}{\sigma}$

And replacing we got:

$z=\frac{38.6- 45}{16}= -0.4$

$z=\frac{57.8- 45}{16}= 0.8$

We can find the probability of interest using the normal standard table and with the following difference:

$P(-0.4$

Step-by-step explanation:

Let X the random variable who represent the expense and we assume the following parameters:

$\mu = 45, \sigma 16$

And for this case we want to find the percent of his expense between 38.6 and 57.8 so we want this probability:

$P(38.6$

And we can assume a normal distribution and then we can solve the problem with the z score formula given by:

$z=\frac{X -\mu}{\sigma}$

And replacing we got:

$z=\frac{38.6- 45}{16}= -0.4$

$z=\frac{57.8- 45}{16}= 0.8$

We can find the probability of interest using the normal standard table and with the following difference:

$P(-0.4$

6 0

3 years ago

I need help with numbers 5,6,and 8

erastovalidia [21]

5) Obtuse triangle because the top angle is more than 90 degrees

6 0

2 years ago

Someone please help me with this lol… have no idea what I’m doing

Sholpan [36]

Given:

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

To find:

The quadrant of the terminal side of $\theta$ and find the value of $\sin\theta$ .

Solution:

We know that,

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II: Only sin and cosec are positive.

In Quadrant III: Only tan and cot are positive.

In Quadrant IV: Only cos and sec are positive.

It is given that,

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

Here cos is positive and sine is negative. So, $\theta$ must be lies in Quadrant IV.

We know that,

$\sin^2\theta +\cos^2\theta =1$

$\sin^2\theta=1-\cos^2\theta$

$\sin \theta=\pm \sqrt{1-\cos^2\theta}$

It is only negative because $\theta$ lies in Quadrant IV. So,

$\sin \theta=-\sqrt{1-\cos^2\theta}$

After substituting $\cos \theta =\dfrac{3}{5}$ , we get

$\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}$

$\sin \theta=-\sqrt{1-\dfrac{9}{25}}$

$\sin \theta=-\sqrt{\dfrac{25-9}{25}}$

$\sin \theta=-\sqrt{\dfrac{16}{25}}$

$\sin \theta=-\dfrac{4}{5}$

Therefore, the correct option is B.

6 0

2 years ago

The position of an object moving along an x axis is given by x = 3.24 t - 4.20 t2 + 1.07 t3, where x is in meters and t in secon

Pavel [41]

Answer and explanation:

Given : The position of an object moving along an x axis is given by $x=3.24t-4.20t^2+1.07t^3$ where x is in meters and t in seconds.

To find : The position of the object at the following values of t :

a) At t= 1 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(1)=3.24(1)-4.20(1)^2+1.07(1)^3$

$x(1)=3.24-4.20+1.07$

$x(1)=0.11$

b) At t= 2 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(2)=3.24(2)-4.20(2)^2+1.07(2)^3$

$x(2)=6.48-16.8+8.56$

$x(2)=-1.76$

c) At t= 3 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(3)=3.24(3)-4.20(3)^2+1.07(3)^3$

$x(3)=9.72-37.8+28.89$

$x(3)=0.81$

d) At t= 4 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(4)=3.24(4)-4.20(4)^2+1.07(4)^3$

$x(4)=12.96-67.2+68.48$

$x(4)=14.24$

(e) What is the object's displacement between t = 0 and t = 4 s?

At t=0, x(0)=0

At t=4, x(4)=14.24

The displacement is given by,

$\triangle x=x(4)-x(0)$

$\triangle x=14.24-0$

$\triangle x=14.24$

(f) What is its average velocity from t = 2 s to t = 4 s?

At t=2, x(2)=-1.76

At t=4, x(4)=14.24

The average velocity is given by,

$\triangle x=x(4)-x(2)$

$\triangle x=14.24-(-1.76)$

$\triangle x=14.24+1.76$

$\triangle x=16$

4 0

3 years ago

Go Math plz try your best

rosijanka [135]

108÷6 = 18 and 84÷6=14 last one is 184÷8=23

4 0

2 years ago