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3 years ago

What is the solution to the equation 7(a – 10) = 13 – 2(2a + 3)

Mathematics

2 answers:

IrinaVladis [17]3 years ago

6 0

The solution to 7(a - 10) = 13 - 2(2a + 3) is x = 7.

Hope this Helps!!

vladimir1956 [14]3 years ago

5 0

Answer:

$a=7$

Step-by-step explanation:

we have

$7(a-10)=13-2(2a+3)$

Solve for a

$7a-70=13-4a-6$

Group terms that contain the same variable and move the constants to the other side

$7a+4a=13-6+70$

Combine like terms

$11a=77$

Divide by $11$ both sides

$a=7$

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Solve the following system using either linear combination or substitution. Round any answers to the hundredths place. Show all

12345 [234]

So if y=-5x-23 then we can plug that value in for y in x-10y=6, so it becomes

x-10(-5x-23)=6

x+50x+230=6

51x=-224

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plug this x value into the equation y=-5x-23

y=-5(-4.39)-23

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3 0

3 years ago

Can anyone please help with this question?

vitfil [10]

The middle section is n(n+1)

The top row is n+2

The bottom row is 2n+1

So the rule for the total number of tiles will be the sum of the above rules...

s(n)=n(n+1)+n+2+2n+1

s(n)=n^2+n+n+2+2n+1

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6 0

3 years ago

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eduard

The answer is 3.7a+3b

Because you combine like terms

3 0

3 years ago

Read 2 more answers

This figure shows △XYZ . MZ¯¯¯¯¯¯ is the angle bisector of ∠YZX . What is XM ? Enter your answer, as a decimal, in the box. unit

astraxan [27]

Answer:

Length of XM is 5.5 units.

Step-by-step explanation:

Given △XYZ where MZ is the angle bisector of ∠YZX . we have to find the length of XM.

A triangle with vertices X, Y, and Z. Side XZ is base. A line segment drawn from Z to M bisects ∠YZX into two parts ∠YZM and ∠XZM.

YZ=7 units, XZ=11 units and YM=3.5 units

By angle bisector theorem which states that an angle bisector of an angle divides the opposite side in two segments that are proportional to the another two sides of the triangle.

Hence, $\frac{YM}{MX}=\frac{YZ}{XZ}$

⇒ $\frac{3.5}{MX}=\frac{7}{11}$

⇒ MX=5.5 units.

Hence, length of XM is 5.5 units.

4 0

3 years ago

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show that thw roots of the equation (x-a)(x_b)=k^2 are always real if a,b and k are real. Please I really need help with this

VLD [36.1K]

Answer:

see explanation

Step-by-step explanation:

Check the value of the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 then roots are real

• If b² - 4ac = 0 roots are real and equal

• If b² - 4ac < 0 then roots are not real

given (x - a)(x - b) = k² ( expand factors )

x² - bx - ax - k² = 0 ( in standard form )

x² + x(- a - b) - k² = 0

with a = 1, b = (- a - b), c = -k²

b² - 4ac = (- a - b)² + 4k²

For a, b, k ∈ R then (- a - b)² ≥ 0 and 4k² ≥ 0

Hence roots of the equation are always real for a, b, k ∈ R

8 0

3 years ago