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3 years ago

What can be a segment bisector

2 answers:

7 0

Answer:A segment bisector, always passes through the midpoint of the segment and divides a segment in two equal parts. ... Points, lines, segments, and rays are all types of segment bisectors. It will be infinite.

Step-by-step explanation:

Vera_Pavlovna [14]3 years ago

6 0

Answer: A segment bisector, always passes through the midpoint of the segment and divides a segment in two equal parts. Points, lines, segments, and rays are all types of segment bisectors. If either a ray or a line serves as a segment bisector, it will be infinite.

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What's the answer to this math problem

GarryVolchara [31]

Total number of marbles:
75 + 50 + 100 + 25 = 250

Probability of selecting a red marble:
25/250 (since 25 out of 250 marbles are red)
Since the top and bottom of this fraction are both divisible by 25, we can go ahead and divide them both by 25, simplifying to 1/10.

Probability of selecting a green marble:
50/250
Simplifies to 2/10
Simplifies to 1/5.

Probability of selecting a blue or orange marble
(75+100)/250
175/250
Simplifies to 7/10.

To find the probability that each event occurs in order, simply multiply them all together.

$\frac{1}{10}\times\frac{1}{5}\times\frac{7}{10}=\frac{1\times1\times7}{10\times5\times10}=\boxed{\frac{7}{500}}$

4 0

3 years ago

Read 2 more answers

Country days scholarship rounds receive a gift of $135000. The money is invested in stock, bonds, and CDs. CDs pay 2.75% interes

Vsevolod [243]

Answer:

CDs — $10,000
bonds — $80,000
stocks — $45,000

Step-by-step explanation:

Let the variables c, b, s represent the dollar amounts invested in CDs, stocks, and bonds, respectively. Then the problem statement gives us 3 relations between these 3 variables:

c + b + s = 135000 . . . . . . . . . . . . . . . . . total invested

0.0275c +.045b +0.104s = 8555 . . . . . total income earned

-c + b = 70000 . . . . . . . . . . . . . . . . . . . . . 70,000 more was in bonds than CDs

Using the third equation to write an expression for b, we can substitute into the other two equations.

b = 70000 +c . . . . . . . . . . . . . . . . expression we can substitute for b

c + (70000 +c) +s = 135000 . . . . substitute for b in the first equation

2c +s = 65000 . . . . . . . . . . . . . . . . [eq4] simplify

.0275c +.045(70000 +c) +.104s = 8555 . . . . . substitute for b in 2nd eqn

.0725c +.104s = 5405 . . . . . . . . . . [eq5] simplify

Using [eq4], we can write an expression for s that can be substituted into [eq5].

s = 65000 -2c . . . . . . . expression we can substitute for s

0.0725c +0.104(65000 -2c) = 5405

-0.1355c = -1355 . . . . . . . . . . . . . . . . . . . . subtract 6760, simplify

c = 1355/.1355 = 10,000

s = 65000 -2×10000 = 45,000

b = 70000 +10000 = 80,000

The amounts invested in stocks, bonds, and CDs were $45,000, $80,000, and $10,000, respectively.

_____

Alternatively, you can reduce the augmented matrix for this problem to row-echelon form using any of several calculators or on-line sites. That matrix is ...

$\left[\begin{array}{ccc|c}1&1&1&135000\\0.0275&0.045&0.104&8555\\-1&1&0&70000\end{array}\right]$

6 0

3 years ago

F(x)=2x and g(x)=square root of x what is the domain of f(g(x))?

Dmitriy789 [7]

Answer:

$x\ge 0$

Step-by-step explanation:

The given functions are:

$f(x)=2x$

and

$g(x)=\sqrt{x}$

Now,

$f(g(x))=f(\sqrt{x} )$

$f(g(x))=2\sqrt{x}$

The domain refers to values of $x$ for which $f(g(x))=2\sqrt{x}$ is defined.

The expression under the radical sign must not be negative.

The domain is

$x\ge 0$

5 0

3 years ago

Show work and Factor 4x^2+xy-18y^2

grigory [225]

Answer:

$4x^2+xy-18y^2=(4x+9y)\,(x-2y)$

Step-by-step explanation:

Let's examine the following general product of two binomials with variables x and y in different terms:

$(ax+by)\,(cx+dy)= ac\,x^2+adxy+bcxy+bdy^2$

so we want the following to happen:

$a\,c = 4\\ad+bc=1\\bd--18$

Notice as well that $ad+bc =1$ means that those two products must differ in just one unit so, one of them has to be negative, or three of them negative. Given that the product $bd=-18$ , then we can consider the case in which one of this (b or d) is the negative factor. So let's then assume that $a\,\,and \,\,c$ are positive.