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3 years ago

Which of these equations has no solution? OA) m = 0 OB) m = m = O c) 2m + 1 = m - 2 od m + 4 = m-4

Mathematics

2 answers:

TiliK225 [7]3 years ago

6 0

Answer:

OD is the correct answer

Veronika [31]3 years ago

6 0

Answer:

OD ) m + 4 = m - 4

Step-by-step explanation:

m + 4 = m - 4

m = m - 4 - 4

m = m -8

m -m = - 8

0 ≠ 8

So, this has no solution

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Write 4,670,000 in standard form

GrogVix [38]

Step-by-step explanation:

4,670,000

4.67x10^6

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3 years ago

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How to write an exponential function whose graph passes through the points (0, -6) and (-2, -54).

suter [353]

Work shown above! Answer is y = -6(1/3)^x
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3 years ago

Please help me to prove this!

Sophie [7]

Answer: see proof below

<u>Step-by-step explanation:</u>

Given: A + B = C → A = C - B

→ B = C - A

Use the Double Angle Identity: cos 2A = 2 cos² A - 1

→ (cos 2A + 1)/2 = cos² A

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]

Use Even/Odd Identity: cos (-A) = cos (A)

<u>Proof LHS → RHS:</u>

LHS: cos² A + cos² B + cos² C

$\text{Double Angle:}\qquad \dfrac{\cos 2A+1}{2}+\dfrac{\cos 2B+1}{2}+\cos^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2+\cos 2A+\cos 2B\bigg)+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\cos^2 C$

$\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C$

$\text{Given:}\qquad \qquad 1+\cos C\cdot \cos (A-B)+\cos^2C$

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$\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)$

$\text{Given:}\qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+A}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-B}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1+2\cos C \cdot \cos A\cdot \cos (-B)$

$\text{Even/Odd:}\qquad \qquad 1+2\cos C \cdot \cos A\cdot \cos B\\\\\\.\qquad \qquad \qquad \quad =1+2\cos A \cdot \cos B\cdot \cos C$

LHS = RHS: 1 + 2 cos A · cos B · cos C = 1 + 2 cos A · cos B · cos C $\checkmark$

5 0

3 years ago

The length of a rectangle is 11cm more than the width. The perimeter is 90cm. Find the length and width of the rectangle.

nadya68 [22]

Answer:

Length=22.5cm, Breadth=11.5cm

Step-by-step explanation:

Let length be l and width be w.

w+22=l

w+w+w+22+w+22=90

4w=90-22-22

=46

w=46 divided by 4

=11.5

l=11.5+11

=22.5

Hence,

Length of rectangle=22.5cm

Breadth of rectangle=11.5cm

3 0

3 years ago

Mrs. Schmidt overlooked a bill and

Flauer [41]

Answer:

Step-by-step explanation:

It's besacilly asking what percentage of 90 is 15

15/90 is equal to 1/6

1/6 is about 16.67

8 0

3 years ago

Which of these equations has no solution? OA) m = 0 OB) m = m = O c) 2m + 1 = m - 2 od m + 4 = m-4​

Which of these equations has no solution? OA) m = 0 OB) m = m = O c) 2m + 1 = m - 2 od m + 4 = m-4