All categories

2 years ago

If a point is inside a circle, the distance from the center of the circle to that point ____.

Mathematics

2 answers:

Alenkasestr [34]2 years ago

5 0

Answer:

A. is less than the radius.

Step-by-step explanation:

Tomtit [17]2 years ago

4 0

C. Passes through the center

You might be interested in

A map has a scale of 1m represents 2 km. write this scale as a ratio in its simplest form

Lisa [10]

Answer:

Step-by-step explanation:

First, find yourself a map. Then, using two points, find both the distance on the map and the true distance. Next, you divide the true distance by the measured map distance, and find your scale. Last, you need to place that ratio onto your map.

8 0

2 years ago

Read 2 more answers

Evaluate 5^28·5^-32·5^-4 A. 17,920 B. 0.017920 C. 0.000256 D. 0.00000256

romanna [79]

Answer:

D 0.00000256

Step-by-step explanation:

$5^{28} *5^{-32}*5^{-4}=5^{28-32-4}=5^{-8}=\frac{1}{5^{8}} =0.00000256$

4 0

2 years ago

Read 2 more answers

The diameter of a circle is 46 yards. What is the circle's circumference? d=46 yd Use 3.14 for n.

kolbaska11 [484]

Answer:

144.44 yards

Step-by-step explanation:

The formula to find circumference is: C = π · d

Since we already have most of the factors figured out, we just replace them with what we have.

C = 3.14 · 46 yrds

144.44 = 3.14 · 46

5 0

2 years ago

Hey guys im new here please solve this for me with steps! ill mark as the best answer

Vinil7 [7]

Answer:

The factors of $2(x+y)^2-9(x+y)-5$ is ((x+y)-5)(2x+2y+1)

Step-by-step explanation:

Given polynomial

=> $2(x+y)^2-9(x+y)-5$

To Find:

The factors of the polynomial =?

Solution:

Lets assume k = (x+y)

Then $2(x+y)^2-9(x+y)-5$ can be written as $2k^2-9k-5$

Now by using quadratic formula

k = $\frac{-b\pm\sqrt{(b^2-4ac}}{2a}$

where

a= 2

b= -9

c= -5

Substituting the values, we get

k = $\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}$

k = $\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}$

k = $\frac{-(-9) \pm \sqrt{(81+40)}}{4}$

k = $\frac{-(-9) \pm \sqrt{(121)}}{4}$

k = $\frac{-(-9) \pm 11}}{4}$

k= $\frac{ 9 \pm 11}{4}$

k = $\frac{20}{4}$ k = $\frac{-2}{4}$

$k_1 =5$ $k_2 = -\frac{1}{2}$

$2k^2-9k-5= 2(k-5)(k+\frac{1}{2})$

Solving the RHS we get

$\frac{2}{2}(k-5)(2k+1)$

$(k-5)(2k+1)$

Substituting k = x+y

$((x+y)-5)(2(x+y+1)$

$((x+y)-5)(2x+2y+1)$

5 0

2 years ago

Express 15 kobo as a decimal of #3.00

arsen [322]

Answer:

15 kobocoin =30.143471

5 0

1 year ago