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3 years ago

The nth term of a sequence is 2n-1.

Mathematics

1 answer:

damaskus [11]3 years ago

5 0

Answer: 25, 31 and 37

Step-by-step explanation:

(2n-1)= 19, 21, 23, 25, 27, 29, 31, 33, 35, 37

(3n+1)= 19, 22, 25, 28, 31, 34, 37,

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HELP ASAP!!

Tresset [83]

By definition, two angles are supplementary if the sum of them is 180 degrees. In this case (see figure attached with the answer) the line AD is transversal to lines AB and DC. This is a proof of the Same-side interior angle theorem.

This theorem states that if we have two lines that are parallel and we intercept those two lines with a line that is transversal to both, same-side interior angles are formed, and also sum 180º, in other words, they are supplementary angles.

Then:

By the definition of a parallelogram, AB∥DC. AD is a transversal between these sides, so ∠A and ∠D are same-side interior angles. Because AB and DC are parallel, the same-side interior angles must be supplementary by the same-side interior angles theorem. Therefore, ∠A and ∠D are supplementary.

3 0

3 years ago

Read 2 more answers

Find the area of the figure. If necessary, round to the nearest tenth.

nirvana33 [79]

Answer:9.4cm2

Step-by-step explanation:

8 0

3 years ago

someone please explain to me how to find the area PLEASE! Find the area and perimeter of quadrilateral ABCD below. Explain your

Lunna [17]

Answer:

Part A) The area of the figure is $24\ units^{2}$

Part B) The perimeter of the figure is $20\ units$

Step-by-step explanation:

step 1

Find the area of the figure

we know that

The area of the figure is equal to the area of triangle ABD plus the area of triangle BCD

The area of triangle is equal to

$A=\frac{1}{2}bh$

Area of triangle ABD

Observing the graph

$b=BD=(-2+8)=6\ units$

$h=(9-5)=4\ units$

substitute

$A=\frac{1}{2}(6)(4)=12\ units^{2}$

Area of triangle BCD

Observing the graph

$b=BD=(-2+8)=6\ units$

$h=(5-1)=4\ units$

substitute

$A=\frac{1}{2}(6)(4)=12\ units^{2}$

The area of the figure is

$12\ units^{2}+12\ units^{2}=24\ units^{2}$

step 2

Find the perimeter of the figure

we know that

The perimeter of the figure is equal to

$P=AB+BC+CD+AD$

we have

$A(-5,1),B(-8,5),C(-5,9),D(-2,5)$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

$d=\sqrt{(5-1)^{2}+(-8+5)^{2}}=5\ units$

Find the distance BC

$d=\sqrt{(9-5)^{2}+(-5+8)^{2}}=5\ units$

Find the distance CD

$d=\sqrt{(5-9)^{2}+(-2+5)^{2}}=5\ units$

Find the distance AD

$d=\sqrt{(5-1)^{2}+(-2+5)^{2}}=5\ units$

substitute the values

$P=5+5+5+5=20\ units$

7 0

3 years ago

Sin5A-sin7A-sin4A+sin8A/cos4A-cos5A-cos8A +cos7A

Anna11 [10]

Sin1A/-cos1A=iddkkkk

5 0

3 years ago

A(1)=20 a(n)=a(n−1)−17 Find the 3rd term in the sequence

Alexxx [7]

Answer: The third term is -14

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

Explanation:

First term = A(1) = 20, which is given

Second term
A(n) = A(n-1) - 17
A(2) = A(2-1) - 17
A(2) = A(1) - 17 .... note: second term = (first term) - 17
A(2) = 20 - 17
A(2) = 3

Third term
A(n) = A(n-1) - 17
A(3) = A(3-1) - 17
A(3) = A(2) - 17 ... note: third term = (second term) - 17
A(3) = 3 - 17
A(3) = -14

6 0

3 years ago