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3 years ago

13

The length of a picture frame is 7 inches more than the width. For what values of x is the perimeter of the picture frame greate

r than 154 inches?

1 answer:

makkiz [27]3 years ago

4 0

P=2(L+W)
l=x+7
w=x
P=2(x+7+x)
P=2(2x+7)
P=4x+14

hast to be greater than 154

P>154
4x+14>154
minus 14 both sides
4x>140
divide 4
x>35

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sandra has $691.43 in her checking account how much does she have in her account after she makes on withdrawal of $327.19 and a

nordsb [41]

Answer:

576.99

Step-by-step explanation:hope this helps;)

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3 years ago

Sin tita= 0.6892.find the value Of tita correct to two decimal places<br><br>

Anvisha [2.4K]

Answer:

$\theta \approx 6.28n + 2.38, \quad n \in \mathbb{Z}$

or

$\theta \approx 6.28n + 0.76, \quad n \in \mathbb{Z}$

Considering $\theta \in (0, 2\pi]$

$\theta \approx 2.38$

or

$\theta \approx 0.76$

Step-by-step explanation:

$\sin(\theta)=0.6892$

We have:

$\sin (x)=a \Longrightarrow x=\arcsin (a)+2\pi n \text{ or } x=\pi -\arcsin (a)+2\pi n \text{ as } n\in \mathbb{Z}$

Therefore,

$\theta= \arcsin (0.6892)+2\pi n, \quad n \in \mathbb{Z}$

or

$\theta = \pi -\arcsin (0.6892)+2\pi n, \quad n\in \mathbb{Z}$

---------------------------------

$\theta \approx 6.28n + 2.38, \quad n \in \mathbb{Z}$

or

$\theta \approx 6.28n + 0.76, \quad n \in \mathbb{Z}$

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3 years ago

A particle moves on the hyperbola xy=18 for time t≥0 seconds. At a certain instant, y=6 and dydt=8. What is x that this instant?

professor190 [17]

Answer:

The value of x at this instant is 3.

Step-by-step explanation:

Let $x\cdot y = 18$ , we get an additional equation by implicit differentiation:

$x\cdot \frac{dy}{dt}+y\cdot \frac{dx}{dt} = 0$ (1)

From the first equation we find that:

$x = \frac{18}{y}$ (2)

By applying (2) in (1), we get the resulting expression:

$\frac{18}{y}\cdot \frac{dy}{dt}+y\cdot \frac{dx}{dt} = 0$ (3)

$y\cdot \frac{dx}{dt}=-\frac{18}{y}\cdot \frac{dy}{dt}$

$\frac{dx}{dt} = -\frac{18}{y^{2}} \cdot \frac{dy}{dt}$

If we know that $y = 6$ and $\frac{dy}{dt} = 8$ , then the first derivative of x in time is:

$\frac{dx}{dt} = -\frac{18}{6^{2}} \cdot (8)$

$\frac{dx}{dt} = -4$

From (1) we determine the value of x at this instant:

$x\cdot \frac{dy}{dt} = -y\cdot \frac{dx}{dt}$

$x = -y\cdot \left(\frac{\frac{dx}{dt} }{\frac{dy}{dt} } \right)$

$x = -6\cdot \left(\frac{-4}{8} \right)$

$x = 3$

The value of x at this instant is 3.

4 0

2 years ago

SOMEONE PLEASE HELP ME! I have a few of the explanations but I'm not sure if I need more please suggest some thank you

lesantik [10]

Answer:

For tingle #1

We can find angle C using the triangle sum theorem: the three interior angles of any triangle add up to 180 degrees. Since we know the measures of angles A and B, we can find C.

$C=180-(A+B)$

$C=180-(21.24+27.14)$

$C=131.62$

We cannot find any of the sides. Since there is noting to show us size, there is simply just not enough information; we need at least one side to use the rule of sines and find the other ones. Also, since there is nothing showing us size, each side can have more than one value.

For triangle #2

In this one, we can find everything and there is one one value for each.

- We can find side c

Since we have a right triangle, we can find side c using the Pythagorean theorem

$b^2=a^2+c^2$

$4^2=2^2+c^2$

$16=4+c^2$

$12=c^2$

$c=\sqrt{12}$

$c=2\sqrt{3}$

- We can find angle C using the cosine trig identity

$cos(C)=\frac{adjacent}{hypotenuse}$

$cos(C)=\frac{2}{4}$

$C=arccos(\frac{2}{4} )$

$C=60$

- Now we can find angle A using the triangle sum theorem

$A=180-(B+C)$

$A=180-(90+60)$

$A=30$

For triangle #3

Again, we can find everything and there is one one value for each.

- We can find angle A using the triangle sum theorem

$A=180-(B+C)$

$A=180-(90+34.88)$

$A=55.12$

- We can find side a using the tangent trig identity

$tan(C)=\frac{opposite-side}{adjacent-side}$

$tan(34.88)=\frac{7}{a}$

$a=\frac{7}{tan(34.88)}$

$a=10.04$

- Now we can find side b using the Pythagorean theorem

$b^2=a^2+c^2$

$b^2=10.04^2+7^2$

$b^2=149.8$

$b=\sqrt{149.8}$

5 0

3 years ago

Helpppppp me plssssss

MaRussiya [10]

I have solved the answer in the pic below, hope it helpss!!!!

8 0

3 years ago