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3 years ago

If PQR measures 75°, what is the measure of SQR? 22° 45° 53° 97°

Mathematics

2 answers:

erastova [34]3 years ago

8 0

Answer:

Step-by-step explanation:

allochka39001 [22]3 years ago

4 0

C is the right answer

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Whats 6 X 8???????????????????????

kifflom [539]

6 x 8 = 48.

(i'm supposed to write longer than 20 characters so here you go)

8 0

3 years ago

Read 2 more answers

A girl and a puppy are 540 feet apart when they start running towards each other. The girl runs at a speed of 15 feet per second

Lerok [7]

Answer: 20 seconds

Step-by-step explanation:

The distance is 540 feet

speed of the girl = 15ft/s

speed of the puppy = 12 ft/s

If we stand in the frame of reference where the girl is still, in this frame the speed of the puppy will be equal to the speed of the puppy in our previus frame plus the speed of the girl in the previous frame (because they are running in opposite directions)

S = 15 ft/s + 12ft/s = 27ft/s

Now we can use the equation:

time = distance/speed

T = 540ft/27ft/s = 20s

6 0

2 years ago

Read 2 more answers

Suppose the graph of y= x^2 translated left 2 units, and up 8 units

Elis [28]

Start with y = x2

Shift 2 units left to get y = (x+2)2

Vertically stretch to get y = 6(x+2)2

Reflect across x-axis to get y = -6(x+2)2

Shift 7 units down to get y = -6(x+2)2 - 7

hope this helps!

3 0

2 years ago

Please help square root math

pantera1 [17]

Answer:

2<∛13<3 1<∛6<2

Step-by-step explanation:

Lets find the integers a and b ∛13 lies between.

a<∛13<b

a³<13<b³

=> a=2 so 2³=8 and b=3 so 3³=27 8<13<27

Similarly a<∛6<b

a³<6<b³

a=1 and b=2

7 0

3 years ago

HELP!! Algebra help!! Will give stars thank u so much <333

Anna35 [415]

Answers:

Part a) $\bf{\sqrt{x^2+(x^2-3)^2}$
Part b) 3
Part c) 2.24
Part d) 1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\$

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying $d = \sqrt{x^2 + (x^2-3)^2}$ is the same as writing d = sqrt(x^2-(x^2-3)^2)

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

Part (b)

Plug in x = 0 and you should find the following

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\$

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

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

Part (c)

Repeat for x = 1

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\$

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

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0

2 years ago