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Allisa [31]
3 years ago
10

Please help me with this question fast!!!!

Mathematics
1 answer:
zubka84 [21]3 years ago
7 0
Reflection and translation
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PLEASE HELP<br><br> Write the number in standard form.<br><br> (8 × 10) + (1 × 110) + (6 × 11,000)
mote1985 [20]
66,190 is the Answer if you add 80 to 110 and then add 190 to 66000
7 0
3 years ago
Estimate the line of best fit using two points on the line.
mash [69]

Answer:

B. y = -2/3x + 12

Step-by-step explanation:

Formula to find the slope when given two points on a line:

<u>y</u><u>2</u><u> </u><u>-</u><u> </u><u>y</u><u>1</u>

x2 - x1

Substitute the two given points (6, 8) (9, 6):

<u>6</u><u> </u><u>-</u><u> </u><u>8</u>

9 - 6

Slope = -2/3x

We found the slope! And the answer choices already gave us one y-intercept, which is 12. The last thing we do is we form an equation with the information we solved and that was given to us.

y = slope (x) + y-intercept

y = -2/3x + 12

The answer choice that matches this equation is B.

In conclusion, the equation that best estimates the line of best fit shown above is answer choice B.

5 0
3 years ago
Read 2 more answers
1. an alloy contains zinc and copper in the ratio of 7:9 find weight of copper of it had 31.5 kgs of zinc.
m_a_m_a [10]

Answer:

Step-by-step explanation:

Question (1). An alloy contains zinc and copper in the ratio of 7 : 9.

If the weight of an alloy = x kgs

Then weight of copper = \frac{9}{7+9}\times (x)

                                      = \frac{9}{16}\times (x)

And the weight of zinc = \frac{7}{7+9}\times (x)

                                      = \frac{7}{16}\times (x)

If the weight of zinc = 31.5 kg

31.5 = \frac{7}{16}\times (x)

x = \frac{16\times 31.5}{7}

x = 72 kgs

Therefore, weight of copper = \frac{9}{16}\times (72)

                                               = 40.5 kgs

2). i). 2 : 3 = \frac{2}{3}

        4 : 5 = \frac{4}{5}

Now we will equalize the denominators of each fraction to compare the ratios.

\frac{2}{3}\times \frac{5}{5} = \frac{10}{15}

\frac{4}{5}\times \frac{3}{3}=\frac{12}{15}

Since, \frac{12}{15}>\frac{10}{15}

Therefore, 4 : 5 > 2 : 3

ii). 11 : 19 = \frac{11}{19}

    19 : 21 = \frac{19}{21}

By equalizing denominators of the given fractions,

\frac{11}{19}\times \frac{21}{21}=\frac{231}{399}

And \frac{19}{21}\times \frac{19}{19}=\frac{361}{399}

Since, \frac{361}{399}>\frac{231}{399}

Therefore, 19 : 21 > 11 : 19

iii). \frac{1}{2}:\frac{1}{3}=\frac{1}{2}\times \frac{3}{1}

             =\frac{3}{2}

     \frac{1}{3}:\frac{1}{4}=\frac{1}{3}\times \frac{4}{1}

              = \frac{4}{3}

Now we equalize the denominators of the fractions,

\frac{3}{2}\times \frac{3}{3}=\frac{9}{6}

And \frac{4}{3}\times \frac{2}{2}=\frac{8}{6}

Since \frac{9}{6}>\frac{8}{6}

Therefore, \frac{1}{2}:\frac{1}{3}>\frac{1}{3}:\frac{1}{4} will be the answer.

IV). 1\frac{1}{5}:1\frac{1}{3}=\frac{6}{5}:\frac{4}{3}

                  =\frac{6}{5}\times \frac{3}{4}

                  =\frac{18}{20}

                  =\frac{9}{10}

Similarly, \frac{2}{5}:\frac{3}{2}=\frac{2}{5}\times \frac{2}{3}

                       =\frac{4}{15}                  

By equalizing the denominators,

\frac{9}{10}\times \frac{30}{30}=\frac{270}{300}

Similarly, \frac{4}{15}\times \frac{20}{20}=\frac{80}{300}

Since \frac{270}{300}>\frac{80}{300}

Therefore, 1\frac{1}{5}:1\frac{1}{3}>\frac{2}{5}:\frac{3}{2}

V). If a : b = 6 : 5

     \frac{a}{b}=\frac{6}{5}

        =\frac{6}{5}\times \frac{2}{2}

        =\frac{12}{10}

  And b : c = 10 : 9

  \frac{b}{c}=\frac{10}{9}

 Since a : b = 12 : 10

 And b : c = 10 : 9

 Since b = 10 is common in both the ratios,

 Therefore, combined form of the ratios will be,

 a : b : c = 12 : 10 : 9

7 0
3 years ago
Make triangles and find the central angle
lubasha [3.4K]

Answer:

Step-by-step explanation:

It 14

3 0
3 years ago
Read 2 more answers
Does anyone know the answer to this one?
polet [3.4K]

Answer:

\large\boxed{a(x+4)(x-5)=0,\ a\in\mathbb{R}-\{0\}}\\\\\text{for}\ a=1\to\boxed{x^2-x-20=0}

Step-by-step explanation:

\text{If}\ x_1\ \text{and}\ x_2\ \text{are the roots of the quadratic equation}\ ax^2+bx+c=0,\\\text{then}\ ax^2+bx+c=a(x-x_1)(x-x_2).\\\\\text{If}\ x_1\ \text{and}\ x_2\ \text{, then they are the roots of the quadratic equation.}\\\\\text{We have}\ x_1=-4\ \text{and}\ x_2=5.\ \text{Therefore we have the equation:}\\\\a(x-(-4))(x-5)=0\\\\a(x+4)(x-5)=0\qquad\text{for any value of}\ a\ \text{except 0}.

7 0
3 years ago
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