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2 years ago

Solve for x please. First correct answer I’ll mark as brainliest

Mathematics

1 answer:

katrin [286]2 years ago

6 0

Answer:

x=7

Step-by-step explanation:

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The first term of an Ap are x, 2x +1, and 5x+1 Find x and the sum of the l0 terms first

bazaltina [42]

Answer:

145/2

Step-by-step explanation:

d=[2x+1-x] = [5x+1 -(2x+1)]

x+1=3x

x-3x=-1

-2x=-1

x=1/2

therefore first term is 1/2

second term is 2(1/2)+1= 2

common difference d is 2-1/2=3/2

sum of first n terms is n/2 [2a+(n-1)d] where n=10

sum of first 10 terms= 10/2[2×1/2+(10-1)3/2]

=5[1+(9)3/2] =5[1+27/2] = 5[29/2] =145/2

7 0

3 years ago

kupik [55]

Using inductive reasoning, the best answer from the given shapes is:
B) In a parallelogram, consecutive angles are supplementary.

7 0

3 years ago

Find the area of the triangle given below.

Nimfa-mama [501]

Given:

A triangle with a base length of 22 ft and a height of 20 ft.

To find:

The area of the given triangle.

Solution:

The area of a triangle is obtained by multiplying $\frac{1}{2}$ with its base length and its height.

The area of a triangle $= \frac{1}{2} (b)(h).$

The height is the distance from the base to the highest point on the triangle and the length of the base is the base length.

The given triangle has a base length of 22 ft and a height of 20 ft.

The area of the triangle $= \frac{1}{2} (b)(h) = \frac{1}{2} (22)(20).$

$\frac{1}{2} (22)(20) = \frac{1}{2} (440) = 220$ ft².

The given triangle has an area of 220 ft².

5 0

3 years ago

Can someone help me with this

antiseptic1488 [7]

Your answer is A

24.6

5 0

3 years ago

The area of the base of a rectangular prisim four and three fourths and the height of two and one third what is the volume

VikaD [51]

Data:
$A_{base} = 4 \frac{3}{4} = \frac{4*4+3}{4} = \frac{16+3}{4} \to A_{base} = \frac{19}{4}$

$h (height) = 2 \frac{1}{3} = \frac{3*2+1}{3} = \frac{6+1}{3} \to h (height) = \frac{7}{3}$
$v (volume) = ?$

Formula: $V = A_{b}*h$

Solving:
$V = A_{b}*h$
$V = \frac{19}{4} * \frac{7}{3}$
$V = \frac{133}{12}$
$V = 11.08333....$
$\boxed{\boxed{V \approx 11.08}}\end{array}}\qquad\quad\checkmark$

7 0

3 years ago