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2 years ago

Could someone help me understand this problem?

Mathematics

2 answers:

Morgarella [4.7K]2 years ago

7 0

Answer:

pick the answer that explains the graph

Step-by-step explanation:

hope this helps a bit??

Julli [10]2 years ago

5 0

Pick the answer that explains the graph

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You choose two different songs on a music playlist at random. Out of 80 songs on the playlist, 36 are hip hop songs. The first s

dexar [7]

Answer:

35/79

Step-by-step explanation:

Assuming you can't choose the same song a second time, the probability is 35/79.

If you can choose the same song, the probability is 36/80, which reduces to 9/20.

7 0

3 years ago

Mathematics Areas and perimeters

sweet-ann [11.9K]

Answer:

54.5 Sq. cm

Step-by-step explanation:

<em>area of a right angled triangle = 1/ 2 × base × height </em>

(here, the base is 8 and height is 6)

= 1/ 2 × 6 × 8

= 4 × 6

<u>= 24 Sq. cm</u>

<em>area of a circle</em> = $\pi$ r²

(r is the radius that is diameter/ 2, and the value of $\pi$ is 3. 14)

r = 10/ 2 = 5 cm

= 3. 14.× (5)²

= 3.14 × 25

<u>= 78.5 Sq. cm</u>

area of the shaded region = area of Circle - area of triangle

= 78.5 - 24

that is option 3

7 0

2 years ago

Can somebody help me with this question please ?

inysia [295]

Answer: No.

Step-by-step explanation: g should also be factored out. 10g(a - 3z).

4 0

2 years ago

What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

So,

$b=0,b-4=0$

$b=0,b=4$

Therefore, the solutions are:

$b=0,\:b=4$

4 0

3 years ago

Read 2 more answers

How do I mutiply 14(a+5+6b)

padilas [110]

Multiply 14 to each term: 14a+70+84b

4 0

2 years ago

Read 2 more answers