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nataly862011 [7]
3 years ago
8

Solve the equation? 16x^2+4=0 Pls help!

Mathematics
1 answer:
miv72 [106K]3 years ago
8 0

Answer:

There is no real solution


Step-by-step explanation:


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A point moves in the plane in the following sequence.
dem82 [27]

Answer:

D

Step-by-step explanation:

The answer is point D(-2,1).

1. If the point D was moved down 2 units, then its coordinates became (-2,-1).

2. If point (-2,-1) was reflected over the x-axis, then its coordinates became (-2,1).

3. If the point (-2,1) was moved 4 units to the right, then its coordinates became (2,1).

4. If point (2,1) was reflected over y-axis, its coordinates became (-2,1).

4 0
2 years ago
The figure shows a circle with center P and inscribed isosceles Triangle ABC. If AC has the same length as the radius of the cir
Ksenya-84 [330]

Answer:

<ABC = 30^{0} (The central angle of a circle is twice any inscribed angle subtended by the same arc).

Step-by-step explanation:

From the diagram, ABC is an inscribed isosceles triangle. But the radius of the circle equals the length AC.

Join P to A and C to form an equilateral triangle. An equilateral triangle has equal sides and angles. So, the value of each interior angle of the equilateral triangle is;

Sum of angle in a triangle = 180^{0}

So that each interior angle = \frac{180^{0} }{3}

                                            = 60^{0}

The value of each interior angle of the triangle is 60^{0}. Thus, <APC = 60^{0}.

⇒ <ABC = 30^{0} (The central angle of a circle is twice any inscribed angle subtended by the same arc.)

7 0
3 years ago
A crawling tractor sprinkler is located as pictured below, 100 feet South of a sidewalk. Once the water is turned on, the sprink
deff fn [24]

Answer:

  a) see below

  b) 32 minutes after turn-on

  c) 52 minutes after turn-on

  d) 20 minutes

  e) 6856.6 ft²

Step-by-step explanation:

a) We have elected to put the origin at the point where the hose crosses the south edge of the sidewalk. Units are feet. Then the sprinkler starts at (0, -100). After 1 hour, 3600 seconds, the sprinkler is 1800 inches, or 150 ft north of where it started, so stops at (0, 50).

The lines forming the sidewalk boundaries are y=0 and y=10.

__

b) Water will first strike the sidewalk when the sprinkler is 20 feet south of it, or 80 feet north of where it started. The sprinkler travels that distance in ...

  (80 ft)(12 in/ft)/(1/2 in/s)(1 min/(60 s)) = 32 min . . . time to start sprinkling sidewalk

__

c) The sprinkler has to travel to a point 130 ft north of its starting position for the water to fall north of the sidewalk. That distance is traveled in ...

  (130 ft)(2/5 min/ft) = 52 min . . . time until end of sprinkling sidewalk

Note that we have combined the scale factors in the expression of part b into one scale factor of (2/5 min/ft).

__

d) The difference of times in parts b and c is the time water falls on the sidewalk: 20 minutes.

__

e) In one hour, the sprinkler travels a distance of ...

  (60 min)(5/2 ft/min) = 150 ft

Of that distance, 10 feet is sidewalk. So, the sprinkler covers an area of grass that is a 140 ft by 40 ft rectangle and a circle of 20 ft radius. The total area of that is ...

  A = LW + πr² = (140 ft)(40 ft) +π(20 ft)² = (14+π)(400) ft² ≈ 6856.6 ft²

The area of grass watered in 1 hour is about 6856.6 ft².

5 0
3 years ago
The equation 2x^2 – 8x = 5 is rewritten in the form of 2(x – p)^2 + q = 0. what is the value of q?
QveST [7]

Answer:

q = -13

Step-by-step explanation:

The given equation is:

2x^{2}-8x=5

Taking 2 as common from left hand side, we get:

2(x^{2}-4x)=5\\\\2[x^{2} - 2(x)(2)]=5

The square of difference is written as:

(a-b)^{2}=a^2 - 2ab + b^{2}                            Equation 1

If we compare the given equation from previous step to formula in Equation 1, we note that we have square of first term(x), twice the product of 1st term(x) and second term(2) and the square of second term(2) is missing. So in order to complete the square we need to add and subtract square of 2 to right hand side. i.e.

2[x^{2}-2(x)(2)+(2)^2-(2)^{2}]=5\\\\ 2[x^{2}-2(x)(2)+(2)^2]-2(2)^{2}=5\\\\ 2(x-2)^{2}-2(4)=5\\\\ 2(x-2)^2-8=5\\\\ 2(x-2)^{2}-8-5=0\\\\ 2(x-2)^{2}-13=0

Comparing the above equation with the given equation:

2(x-p)^{2}+q=0, we can say:

p = 2 and q= -13

6 0
3 years ago
Please help with this scientific notation thing
12345 [234]

Answer:

Scentific notation means that the coefficient has to be between 1 and 10.

Step-by-step explanation:

If its not, add or subtract that number idrk

7 0
2 years ago
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