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2 years ago

Geometry question!! Please help i’ll give points!

Mathematics

1 answer:

Semenov [28]2 years ago

4 0

Answer:

63°

Step-by-step explanation:

$m\angle OZQ=m\angle OZP+m\angle PZQ \\ \\ 125^{\circ}=62^{\circ}+m\angle PZQ \\ \\ m\angle PZQ=63^{\circ}$

You might be interested in

Which fraction make comparison 3/4

velikii [3]

Answer:

6/8, 12/16, and so on. I wasn't sure how you wanted this question to be answered, but 3/4 is also 0.75

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Pleaseeeeeeee I need help nowwwww

Tanya [424]

Answer:

(2)

Step-by-step explanation:

One property of a parallelogram is

• opposite angles are congruent, hence

∠L = ∠N and ∠M = ∠P

Using ∠L = ∠N, then

8x - 41 = 6x + 3 ( subtract 6x from both sides )

2x - 41 = 3 ( add 41 to both sides )

2x = 44 ( divide both sides by 2 )

x = 22

-------------

∠M = 2x + 1 = (2 × 22) + 1 = 44 + 1 = 45°

∠P = ∠M = 45° → (2)

6 0

4 years ago

Directio: Draw and label a figure for the condition.

fgiga [73]

Answer:

See attached

Step-by-step explanation:

Rays AB and AM intersect at point A

7 0

3 years ago

Find a polynomial of the form

Sloan [31]

Answer:

$f(x)=-\frac{1}{4}x^3-\frac{1}{6}x^2+\frac{65}{12}x-3$

Step-by-step explanation:

We are given that

$f(x)=ax^3+bx^2+cx+d$

f(0)=-3,f(1)=2,f(3)=5 and f(4)=0

We have to find the polynomial

Substitute the value x=0 then ,we get

f(0)=d=-3

Substitute x=1 then we get

$a+b+c-3=2$

$a+b+c=2+3=5$

$a+b+c=5$ (equation I)

Substitute x=3 then we get

$a(3)^3+b(3)^2+c(3)-3=5$

$27a+9b+3c=5+3=8$

$27a+9b+3c=8$ (Equation II)

Substitute x=4 then we get

$a(4)^3+b(4)^2+c(4)-3=0$

$64a+16b+4c=3$ (Equation III)

Equation I multiply by 3 then subtract from equation II

$24a+6b=-7$ (Equation IV)

Equation II multiply by 4 and equation III multiply by 3 and subtract equation II from III

$84a+12b=-23$ (Equation V)

Equation IV multiply by 2 and then subtract from equation V

$36a=-9$

$a=-\frac{9}{36}$

$a=-\frac{1}{4}$

Substitute the value of a in equation IV then we get

$24(-\frac{1}{4})+6b=-7$

$-6+6b=-7$

$6b=-7+6=-1$

$b=-\frac{1}{6}$

Substitute the value of b in equation I then we get

$-\frac{1}{4}-\frac{1}{6}+ c=5$

$-\frac{5}{12}+c=5$

$c=5+\frac{5}{12}=\frac{65}{12}$

Substitute the values then we get

$f(x)=-\frac{1}{4}x^3-\frac{1}{6}x^2+\frac{65}{12}x-3$

5 0

4 years ago

Please help me!!! Will get brainliest!!!

Alex73 [517]

Using the formula
A = Pe^kt
Where A is the ending amount of population, P is the population k is a constant and 't' is time.

A = 0.39, t= 12 and p= 0.29
0.39= 0.29e^12k
1.34 = e^12k
ln1.34 = 12k
ln1.34/12 = k
K= 0.02

No the graph does not support his claim as k should be equal to 0.5

4 0

3 years ago