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nikdorinn [45]
3 years ago
11

Can someone please help me with math.

Mathematics
1 answer:
romanna [79]3 years ago
6 0
Answer:

I think it’s A
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Whats is the slope of line that is perpendicular to the line y= -9x + 7
11Alexandr11 [23.1K]

Answer: 1/9

Step-by-step explanation:

The slope of this line is -9. The slope of the perpendicular line is the opposite reciprocal. The reciprocal of -9 is -1/9, the opposite of that is 1/9

3 0
3 years ago
If you translate the coordinates (1,2) to the right 1 and up 3, what would be the
Svetlanka [38]

Answer:

(2,5)

Step-by-step explanation:

1 to the right is +1

3 up is +3

1+1=2

2+3=5

(2,5)

3 0
3 years ago
To estimate the number of lizards in a certain region, Sean traps 36 lizards and marks them. He releases these lizards and then
nexus9112 [7]

Answer:

The BEST estimate for the lizard population is 450

Step-by-step explanation:

Let

x -----> the estimate for the lizard population

we know that

using proportion

\frac{200}{16}=\frac{x}{36}\\\\x=200*36/16\\\\x=450

therefore

The BEST estimate for the lizard population is 450

8 0
4 years ago
Read 2 more answers
Suppose that $p$ and $q$ are positive numbers for which \[\log_9 p = \log_{12} q = \log_{16} (p + q).\] what is the value of $q/
SVEN [57.7K]

Given p,q>0 and \log_9p=\log_{12}q=\log_{16}(p+q)=x (say).

Then,

p=9^x\\ q=12^x\\ p+q=16^x

From the above 3 equations,

\frac{q}{p} =(\frac{12}{9} )^x\\ \frac{q}{p} =(\frac{4}{3} )^x\\ \frac{p+q}{p} =(\frac{16}{9} )^x\\ \frac{p+q}{p} =(\frac{4}{3} )^{2x}\\

From the equations, we get

\frac{p+q}{p}=(\frac{q}{p})^2\\ 1+\frac{q}{p}=(\frac{q}{p})^2\\ (\frac{q}{p})^2-\frac{q}{p}-1=0\\ \frac{q}{p}=\frac{1 \pm \sqrt{5}}{2}

Since p,q>0, the negative value is rejected.

\frac{q}{p}=\frac{1 + \sqrt{5}}{2}

3 0
4 years ago
Calcola el 28% de 375
Zina [86]

Answer:

Suponiendo que te refieres a calcular el 28% de 375, tu respuesta es 105.

Step-by-step explanation:

\frac{y}{375} :\frac{28}{100}

y · 100 = 28 · 375

100y = 10500

100y ÷ 100 = 10500 ÷ 100

y = 105

5 0
3 years ago
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