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3 years ago

7

PICTURE PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

1 answer:

mariarad [96]3 years ago

4 0

Answer:

Step-by-step explanation:

Because WY⊥YZ and WY⊥WV ⇒ WV║YZ ⇒ ∠YZX ≅ ∠WVX ⇒ ΔXWV ~ ΔXYZ (AAA similarity theorem)

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17 points: Answer the question in the picture provided. Please hurry.

KatRina [158]

PEMDAS
yo might be tricked to do 2(-7) but that's multiplcation

so expoments
(-6)^2=36

now we have
36/12-2(-7)
mulitipictioan or division whicever comes first
division
36/12=3
multiplcation 2(-7)=-14
now we have
3-(-14)
3+14
17
answer is 17

8 0

3 years ago

Can Someone Help Me With This?? :)

kykrilka [37]

Rational) No
Integer) No
Whole) Yes
Natural) Yes
Describe the number:
It’s whole and Natural and it is 5

5 0

4 years ago

I need help with this question fast! i don't have a lot of time! can someone help me please???

jarptica [38.1K]

16x^4-81y^4 difference of perfect squares...

(4x^2-9y^2)(4x^2+9y^2) that's one of the equivalents...

(2x+3y)(2x-3y)(4x^2+9y^2) that's another one of them...

5 0

3 years ago

A collection of nickels and quarters is worth 2.85 . There are 3 more nickels than quarters how many nickels and quarters are th

Zolol [24]

A nickel is equal to 5 cents or 0.05 dollars.

A quarter is equal to 25 cents or 0.25 dollars.

Let number of nickels be = n

Let number of quarters be = q

$0.05n+0.25q=2.85$ ...........(1)

As it is given, there are 3 more nickels than quarters so equation becomes,

$n=q+3$ ................(2)

Plug in the value of 'n' from (2) in (1)

$0.05(q+3)+0.25q=2.85$

= $0.05q+0.15+0.25q=2.85$

$0.30q=2.70$

$q=9$

As $n=q+3$ we get, $n=9+3=12$

Hence, there are 12 nickels and 9 quarters.

3 0

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

3 years ago