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lina2011 [118]
3 years ago
14

Please help me with this question? i’d appreciate it :)

Mathematics
1 answer:
UNO [17]3 years ago
5 0

Answer:

1.2

Step-by-step explanation:

Output of f, which also means f(x), when x = -6 is 1.2.

Using the graph given, this implies that, when x = -6, the corresponding value of f(x) will give us 1.2.

Thus:

f(-6) = 1.2 =>> (-6, 1.2).

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C is the answer

Step-by-step explanation:

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2 years ago
A—15<br> b-30<br> c-36<br> d-72asap
3241004551 [841]

Answer:

C 36

Step-by-step explanation:

1/2(base)(height)

1/2(8)(9)

1/2(72)

=36

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3 years ago
Which option is the best example of a financial emergency?
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ABCD- parallelogram, If the perimeter of Triangle CPQ is 15cm, Find the perimeter of triangle BAQ. Find the perimeter of triangl
melamori03 [73]

Answer:

The answer is below

Step-by-step explanation:

A parallelogram is a quadrilateral (has 4 sides and 4 angle) with two pair of parallel and opposite sides. Opposite sides of a parallelogram are parallel and equal.

Given parallelogram ABCD:

AB = CD = 18 cm; BC = AD = 8 cm

∠P = ∠P, ∠PDA = ∠PCQ (corresponding angles are equal).

Hence ΔPCQ and ΔPDA are similar by angle-angle similarity theorem. For similar triangles, the ratio of their corresponding sides equal. Therefore:

\frac{CD}{PC}= \frac{AD}{CQ}\\\\\frac{18}{6}=\frac{8}{x}  \\\\x=\frac{6*8}{18}=\frac{8}{3}\ cm

Perimeter of CPQ = CP + CQ + PQ

15 = 6 + 8/3 + PQ

PQ = 15 - (6 + 8/3)

PQ = 6.33

∠CQP = ∠AQB (vertical angles), ∠QCP = ∠QBA (alternate angles are equal).

Hence ΔCPQ and ΔABQ are similar by angle-angle similarity theorem

\frac{AQ}{QP}=\frac{AB}{CP}  \\\\\frac{AQ}{6.33} =\frac{18}{6} \\\\AQ=\frac{18}{6}*6.33\\\\AQ = 19

\frac{BQ}{CQ}=\frac{AB}{CP}  \\\\\frac{BQ}{8/3} =\frac{18}{6} \\\\BQ=\frac{18}{6}*\frac{8}{3} \\\\BQ =8

Perimeter of BAQ = AB + BQ + AQ = 18 + 8 + 19 = 45cm

PA = AQ + PQ = 19 + 6.33 = 25.33

PD = CD + DP = 18 + 6 = 24

Perimeter of PDA = PA + PD + AD = 24 + 25.33 + 8 = 57.33 cm

7 0
3 years ago
Find the equation of the line that contains the point (-1,-11) and is parallel to the line 7x+3y=10
madam [21]
Y=- \frac{7}{3} -13 \frac{1}{3}.

To find the equation of a line, you need two things: the slope and the y-intercept. 

The slopes of parallel lines are the same. So we can find the slope of the new line by finding the slope of the first line. To do that, we need to put it in y=mx+b format, where m is the slope. So we must rearrange the 7x+3y=10. First subtract 7x from both sides to make it look like:
       3y=10-7x
Then divide both sides three:
       by= \frac{10}{3} - \frac{7}{3} x
So now that it's in y=mx+b format, we can now see that the m= - \frac{7}{3}

Now we know the m of the new equation, we need to find the b, or the y-intercept. To do this, we can plug the point we have and the m value into the y=mx+b format.
       (-11)=- \frac{-7}{3} (-1) + b
Solving this, we can subtract 7/3 from both sides:
     -11- \frac{7}{3} = b
Therefore, b= -13 \frac{1}{3}

Plugging the m= - \frac{7}{3} and the b= -13 \frac{1}{3} back into the y=mx+b format, your parallel line is y=- \frac{7}{3} -13 \frac{1}{3}.
5 0
3 years ago
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