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3 years ago

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Annie is adding the expression 4 + 5 + (–4) and follows the procedure of starting at 4, moving right 5 units, and then moving le

ft 4 units.
Beth says she knows another way to find the sum, using the additive inverse property. What procedure could Beth use?

1 answer:

jekas [21]3 years ago

4 0

Answer:

Step-by-step explanation:

hello

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Is (0, -1) a solution of -101 + 4y > -4?

Andrei [34K]

Answer:

yes

Step-by-step explanation:

4 0

3 years ago

Rectangle ABCD has vertices at (-7,-2); (1, -2); (1, -8); and (-7, -8) respectively. If GHJK is a similar rectangle where G(2, 5

Minchanka [31]

Answer:

J (6,2) K (2,2)

Step-by-step explanation:

Now distance of AB = [(y2-y1)^2 + (x2-x1)^2]^(1/2)

= 8 units

CB = [(y2-y1)^2 + (x2-x1)^2]^(1/2)

= 6 units by using same formula

Now AB/CB must equal to GH/HJ as rectangles are similar

GH = [(y2-y1)^2 + (x2-x1)^2]^(1/2)

= 4 units

so

8/6 = 4/HJ

So,

HJ = 3 units

Now if we see the coordinates given carefully, it is obvious that two perpendicular lines lie perfectly parallel to x and y coordinates in rectangle ABCD. A is (-7,-2) and B is (1,-2) which means distance along y-axis doesn't change. Similarly for C (1,-8) and D (-7,-8), one can see that distance between y-axis doesn't change. So lines AB and CD of rectangle are parallel with x and AD and BC are parallel with y-axis.

In rectangle GHJK one can see that in given coordinates, G(2,5) and H(6,5), y coordinate is same so it is parallel to x axis. Now, HJ is perpendicular to GH so it must be parallel to y axis. It means if we know the lengths of sides we can easily determine unknown coordinates by simple addition and subtraction.

So, we know HJ = 3 units

J is (6,2) since HJ is parallel to y axis so distance on x axis will remain unchanged and length of line HJ will effect distance of y axis.

Similarly K is (2,2) for the same reason.

6 0

3 years ago

A rectangle has sides 2x + 3 and x - 4. what is the perimeter of the rectangle

amid [387]

Write 2x+3(to the power of 2)+x-4(to the power of 2). Then do the equations and you should get 4x+9+2x+16. Combine like terms and that's your answer. I know it might be weird that it's not like 72 or something, but it's math haha. Your answer should be 6x+25! Make sure you understand this before just writing all of this down. Also, don't be afraid to ask your teacher. They are there to help!

8 0

3 years ago

Tobias's music store bought a piano at the wholesale price of $550. Tobias marked up the price 35%. What was the amount of marku

Mrrafil [7]

Answer:

A. $58.50

B. $80

C. No. The markup was 33 1/3%

Step-by-step explanation:

You are expected to know the relevant relationship is ...

cost + markup = selling price . . . . also called "store price" in this problem

___

A. The markup is 30% × $45 = $13.50, so the store price is ...

$45 + 13.50 = $58.50

___

B. When the 30% markup is added to the 100% cost, the selling price is 130% of the cost, or 1.30 times the cost. Here you have ...

$104 = 1.30 × cost

$104/1.3 = cost = $80 . . . . . divide by the coefficient of cost

___

C. The markup is $100 -75 = $25, so the markup percentage based on cost is ...

$25/$75 × 100% = 33 1/3%

This is not 30%.

Step-by-step explanation:

8 0

3 years ago

Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to chec

klasskru [66]

Answer:

There is a horizontal tangent at (0,-4)

The tangent is vertical at (-2,-3) and (2,-3).

Step-by-step explanation:

The given function is defined parametrically by the equations:

$x=t^3-3t$

and

$y=t^2-4$

The tangent function is given by:

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\implies \frac{dy}{dx}=\frac{2t}{3t^2-3}$

The tangent is vertical at when $\frac{dx}{dt}=0$

$\implies \frac{3t^2-3}{2t}=0$

$\implies 3t^2-3=0$

$\implies 3t^2=3$

$\implies t^2=1$

$\implies t=\pm1$

When t=1,

$x=1^3-3(1)=-2$ and $y=1^2-4=-3$

When t=-1,

$x=(-1)^3-3(-1)=2$ and $y=(-1)^2-4=-3$

The tangent is vertical at (-2,-3) and (2,-3).

The tangent is horizontal, when $\frac{dy}{dx}=0$ or $\frac{dy}{dt}=0$

$\implies 2t=0$

$\implies t=0$

When t=0,

$x=0^3-3(0)=0$ and $y=0^2-4=-4$

There is a horizontal tangent at (0,-4)

5 0

3 years ago