3 years ago

Can someone help me with this problem ASAP please.

Mathematics

1 answer:

harina [27]3 years ago

3 0

None are correct because angle A is 40 degrees

You might be interested in

Two trees are next to each other in a clearing. The first tree is 13 feet tall and casts an 8-foot shadow. The second tree casts

SVEN [57.7K]

The second tree is 59.5 feet
tall. Given Two trees are growing
in a clearing. The first tree

3 0

2 years ago

What equation can relate to this sentence 11 is the quotient of a number y and 6 ?

Natalija [7]

Answer: 11 = y ÷ 6

Step-by-step explanation: since the problem says 11 is that means the equation equals 11 and when it says quotient that means that the equation must include division.

4 0

3 years ago

Mr. Rodriguez is assembling supply kits for his students. He has 24 pencils and 32 erasers. All the kits must have the same numb

lbvjy [14]

Answer:

1. 8 kits

2. 3 pencils and 4 erasers

4 0

3 years ago

Ben works part time in a warehouse . His boss is going to send one employee home early . Of his 56 employees 21 are part time .

qaws [65]

21/56 which reduces to 3/8 ...or if u need a percent, it is 37.5%

7 0

3 years ago

Read 2 more answers

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