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GREYUIT [131]
3 years ago
15

Choose the correct answer.

Mathematics
1 answer:
Ronch [10]3 years ago
6 0
What figure?? Please elaborate
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I want help its urgent!!! <br><br>(iv), (v), (vi) Questions​
Artist 52 [7]

Solution:

iv) Given equation: x^{2} - y^{2} - 4x - 2y + 3

Since we are given two variables, we need to split the original expression into two quadratic expressions as

x^{2} - 4x + 4 - y^{2} - 2y - 1

Factoring x first:

x^{2} - 4x + 4 = x^{2} - 2x - 2x + 4

= x (x - 2) - 2 (x-2)

= (x-2)(x-2)

= (x-2)^{2}

Factoring y now:

-y^{2} - 2y - 1 = -1(y^{2} + 2y + 1)

= -1(y^{2} + y + y + 1)

= -1[ y(y+1) + 1(y+1)]

= -1 [(y+1)(y+1)]

= (y+1)(-y-1)

Therefore, the original expression becomes

(x-2)^{2}+(y+1)(-y-1)\\or \\(x-2)(x-2) + (y+1)(-y-1)

7 0
3 years ago
1. Consider an athlete running a 40-m dash. The position of the athlete is given by , where d is the position in meters and t is
sasho [114]

There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:

Consider an athlete running a 40-m dash. The position of the athlete is given by d(t)=\frac{t^{2}}{6}+4t where d is the position in meters and t is the time elapsed, measured in seconds.

Compute the average velocity of the runner over the intervals:

(a) [1.95, 2.05]

(b) [1.995, 2.005]

(c) [1.9995, 2.0005]

(d) [2, 2.00001]

Answer

(a) 6.00041667m/s

(b) 6.00000417 m/s

(c) 6.00000004 m/s

(d) 6.00001 m/s

The instantaneous velocity of the athlete at t=2s is 6m/s

Step by step Explanation:

In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:

V_{average}=\frac{d(t_{2})-d(t_{1})}{t_{2}-t_{1}}

so let's take the first interval:

(a) [1.95, 2.05]

V_{average}=\frac{d(2.05)-d(1.95)}{2.05-1.95}

we get that:

d(1.95)=\frac{(1.95)^{3}}{6}+4(1.95)=9.0358125

d(2.05)=\frac{(2.05)^{3}}{6}+4(2.05)=9.635854167

so:

V_{average}=\frac{9.6358854167-9.0358125}{2.05-1.95}=6.00041667m/s

(b) [1.995, 2.005]

V_{average}=\frac{d(2.005)-d(1.995)}{2.005-1.995}

we get that:

d(1.995)=\frac{(1.995)^{3}}{6}+4(1.995)=9.30335831

d(2.005)=\frac{(2.005)^{3}}{6}+4(2.005)=9.363335835

so:

V_{average}=\frac{9.363335835-9.30335831}{2.005-1.995}=6.00000417m/s

(c) [1.9995, 2.0005]

V_{average}=\frac{d(2.0005)-d(1.9995)}{2.0005-1.9995}

we get that:

d(1.9995)=\frac{(1.9995)^{3}}{6}+4(1.9995)=9.33033358

d(2.0005)=\frac{(2.0005)^{3}}{6}+4(2.0005)=9.33633358

so:

V_{average}=\frac{9.33633358-9.33033358}{2.0005-1.9995}=6.00000004m/s

(d) [2, 2.00001]

V_{average}=\frac{d(2.00001)-d(2)}{2.00001-2}

we get that:

d(2)=\frac{(2)^{3}}{6}+4(2)=9.33333333

d(2.00001)=\frac{(2.00001)^{3}}{6}+4(2.00001)=9.33339333

so:

V_{average}=\frac{9.33339333-9.33333333}{2.00001-2}=6.00001m/s

Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s

8 0
3 years ago
Write this rate as a unit rate 7/5 over 2/3
Juli2301 [7.4K]

Answer:

no i dont want too

Step-by-step explanation:

HAHAHAHAHAHHAHAAHHAAHAHAHAHAhahahahahahahahahahaahhhahahahahaahahahahahhhhaahhahahaahahhahaahhahahhahahahahahahahahahahahahahahahahahzhahahahahah

6 0
3 years ago
Which expression is equivalent to y · 48?
gtnhenbr [62]

Answer:

None of the above

Step-by-step explanation:

F - 40y

G - 4y

H - 40y

J y+4

6 0
2 years ago
BRAINLIEST IF CORRECT
Kisachek [45]

Answer:

The two values are 100 x 100 x 100 and 1 x 10 to the power of 3

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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