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3 years ago

The sum of two decimals numbers is 5.3 their difference is 1.7 and their product is 6.3 what are the two numbers

Mathematics

2 answers:

Korolek [52]3 years ago

8 0

Answer:

3.5 and 1.8

Step-by-step explanation:

x=y=5.3

x-y=1.7

x=3.5

y=1.8

n200080 [17]3 years ago

3 0

Answer:

3.5 and 1.8

Step-by-step explanation:

x + y = 5.3

x - y = 1.7

2x = 7.0

for x:

x = 7/2

x = 3.5

for y:

3.5 + y = 5.3

y = 5.3 - 3.5

y = 1.8

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SOS if anyone could help that would be appreciated!!!❤️

Anastaziya [24]

Answer:

See below

Step-by-step explanation:

$2e)15-4x=10x+45$

Solve:

$15-4x=10x+45\\-4x-10x=45-15\\-14x=30\\ \boxed {x=-\dfrac{15}{7}}$ $3(x+1)=21$

Check:

$15-4 \left(-\dfrac{15}{7} \right)=10\left(-\dfrac{15}{7} \right)+45$

$15+\dfrac{60}{7} \right)=-\dfrac{150}{7} +45$

$\dfrac{210}{7} =30$

$30=30$

$3a) 3(x+1)=21$

Solve:

$3(x+1)=21\\3x+3=21\\3x=18\\x=6$

$3b) 2+3(x+1)=6x$

$2+3x+3=6x\\5=3x\\$

$x=\dfrac{5}{3}$

$3c) -2(2x-3)=-2$

Solve:

$-2(2x-3)=-2\\-4x+6=-2\\-4x=-8\\x=2$

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3 years ago

50 POINTS/ WILL MARK BRAINLIEST

Nina [5.8K]

Answer:

i think the answer is She replaced both the divisor and the dividend with their reciprocals when changing division to multiplication in step 2.

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3 years ago

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harina [27]

Answer:

96.5606

Step-by-step explanation:

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sammy [17]

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3 years ago

If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Compute the first derivative. By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta$

and so the second derivative is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta$

4 0

3 years ago