Find two numbers that round 15.5 to the nearest tenth

Mathematics

1 answer:

Hoochie [10]4 years ago

5 0

15.49 and 15.48 any number below 15.50 and above 15.45 will round up to 15.5

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A) 1/5 + 3/10 B) 7/8 - 1/4 Work out then simplify

Brrunno [24]

Answer:

1/5=0,2

3/10=0,3

0,2+0.3=0.5

7/8=0.875

1/4=0.25

0.875-0.25=0.6

4 0

3 years ago

Read 2 more answers

WILL GIVE BRAINLIEST! What is the answer for question (iv)? please help!

Leviafan [203]

Step-by-step explanation:

$\triangle ABC \sim \triangle ADE... (given) \\\\\therefore \frac {AC}{AE} = \frac {BC}{DE}.. (csst) \\\\\therefore \frac {5}{5+2x} = \frac {x+3}{4x+1}\\ \\ \therefore \: 5(4x + 1) = (5 + 2x)(x + 3) \\ \therefore \:20x + 5 = 5x + 15 + 2 {x}^{2} + 6x \\ \therefore \:20x + 5 =2 {x}^{2} + 11x + 15 \\ \therefore \:2 {x}^{2} + 11x + 15 -2 0x - 5 = 0\\ \therefore \:2 {x}^{2} - 9x + 10 = 0\\ \therefore \:2 {x}^{2} - 5x - 4x+ 10 = 0\\ \therefore \:x(2x - 5) - 2(2x - 5) = 0 \\ \therefore \:(2x - 5)(x - 2) = 0 \\ 2x - 5 = 0 \: \: or \: \: x - 2 = 0 \\ x = \frac{5}{2} \: \: or \: \: x = 2 \\ x = 2.5 \: \: or \: x = 2 \\ x = 2.5 \: is \: largest \\ \\ \frac{A(\triangle ABC)}{ A(\triangle ADE)} = \frac{AC ^{2} } {AE^{2} } \\ \\ \frac{A(\triangle ABC)}{ A(\triangle ADE)} = \frac{5 ^{2} } {(5 + 2x)^{2} } \\ = \frac{25}{(5 + 2 \times 2.5)^{2} } \\ = \frac{25}{ {10}^{2} } \\ = \frac{25}{100} \\ \\ \frac{A(\triangle ABC)}{ A(\triangle ADE)} = \frac{1}{4} \\$