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artcher [175]
3 years ago
6

In parallelogram ABCD, AD = 4x - 10 and BC = 3x + 2. Find AD.

Mathematics
1 answer:
Helga [31]3 years ago
8 0

Answer:12


Step-by-step explanation:

4x - 10= 3x+2 add 10= 4x = 3x+12 subtract 3x=1x 1x=12 divide 12 by 1=12 should be 12 sorry if I got it wrong...

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If f(x)=6x-9, find f(4)
kipiarov [429]

Answer:

13

Step-by-step explanation:

In order to find the answer, plug in 4 for x.

f(x)=6x-9\\f(4)=6(4)-9\\f(4)=24-9\\f(4)=13

3 0
2 years ago
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Triangle ABC has vertices at A(-4, -2), B(1, 7) and C(8, -2)
Arada [10]

Answer:

a). Area = 54 square units

b). Perimeter = 33.7 units

Step-by-step explanation:

Vertices of the triangle ABC are A(-4, -2), B(1, 7) and C(8, -2).

(a). Area of the triangle ABC = \frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})] (Absolute value)

By substituting the values from the given vertices,

Area = \frac{1}{2}[(-4)(7+2)+(1)(-2+2)+8(-2-7)]

        = \frac{1}{2}[-36+0-72]

        = \frac{1}{2}(-108)

        = (-54) unit²

Therefore, absolute value of the area = 54 square units

(b). Distance between two vertices (a, b) and (c, d)

        d = \sqrt{(a-c)^{2}+(b-d)^2}

     AB = \sqrt{(-4-1)^{2}+(-2-7)^{2}}

           = \sqrt{106}

           = 10.295 units

     BC = \sqrt{(1-8)^2+(7+2)^2}

           = \sqrt{130}

           = 11.402 units

     AC = \sqrt{(-4-8)^2+(-2+2)^2}

           = 12 units

     Perimeter of the triangle = AB + BC + AC = 10.295 + 11.402 + 12

                                                                          = 33.697

                                                                          ≈ 33.7 units

7 0
3 years ago
Need help as soon as possible please
valina [46]

80 - 9.38 = 70.62

41.54(.07) = 2.91

41.54 + 2.91 = 44.45

66(.07) = 4.62

66 + 4.62 = 70.62

With an $80 gift card, Stephanie can buy a sundress as long as the ticketed price is less than or equal to $66.00.

3 0
3 years ago
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Find the slope of the line that passes through the points (2,4) and (6,12)?
Illusion [34]

Answer:

2 hurry!

Step-by-step explanation:

7 0
3 years ago
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Find the sum of the first 20 terms of the arithmetic sequence 4, -4, -12, -20​
Vsevolod [243]

Answer:

The sum of the first 20 terms is -1440.

Step-by-step explanation:

We want to find the sum of the first 20 terms of the arithmetic sequence:

4, -4, -12, -20...

The sum of an arithmetic sequence is given by:

\displaystyle S=\frac{k}{2}(a+x_k)

Where <em>k</em> is the number of terms, <em>a</em> is the initial term, and <em>x</em>_<em>k</em> is the last term.

Since we want to find the sum of the first 20 terms, <em>k</em> = 20.

Our initial term <em>a</em> is 4.

Our last term is also the 20th term as we want the sum of the first 20 terms.

To find the 20th term, we can write an explicit formula for our sequence. The explicit formula for an arithmetic sequence is given by:

x_n=a+d(n-1)

Where <em>a</em> is the initial term, <em>d</em> is the common difference, and <em>n</em> is the <em>n</em>th term.

Our initial term is 4. From the sequence, we can see that our common difference is -8 since each subsequent term is eight less than the previous term. Therefore:

x_n=4-8(n-1)

Then the last or 20th term is:

x_{20}=4-8(20-1)=4-8(19)=-148

Therefore, the sum of the first 20 terms are:

\displaystyle\begin{aligned} S_{20}&=\frac{(20)}{2}\left((4)+(-148))\\&=10(-144) \\&= -1440\end{aligned}

5 0
2 years ago
Read 2 more answers
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