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3 years ago

Help plz show work asap!!!!!!

Mathematics

1 answer:

Pie3 years ago

6 0

Answer:

150 m

Step-by-step explanation:

So the scale of the model is 1 cm = 5 m, and the model is 30 cm long. We can set this up as two fractions, then cross-multiply, like so:

1 cm/30 cm = 5 m/x m

1 * x = x

30 * 5 = 150

Now, set the two products equal to each other. Normally we'd have to simplify, but x is already on its own, so we have our answer:

x = 150 m

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2 ( n - 7 ) = 20 What is n?

Sunny_sXe [5.5K]

Answer:

The answer is n = 17

Step-by-step explanation:

Divide both sides by the numeric factor on the left side, then solve.

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3 years ago

Classify the following triangle. Check all that apply

Stels [109]

Its equilateral ( all sides are equalA0

Also its acute.

D and F

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3 years ago

How do i tell if y=x^3+x^2 is even or odd?

Effectus [21]

If a function is even, then f(-x) = x.

If a function is odd, then f(-x) = -x.

y = x³ + x² → f(x) = x³ + x² → -f(x) = -(x³ + x²) = -x³ - x²

f(-x) = (-x)³ + (-x)² = [(-1)(x)]³ + [(-1)(x)]² = (-1)³x³ + (-1)²x²

= -1x³ + 1x² =-x³ + x²

f(-x) ≠ f(x) and f(-x) ≠ -f(x)

y = x³ + x² is not odd and not even

Answer: neither

8 0

3 years ago

If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.

Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have $S_1=3\times2^{1-1}+2(-1)^1=3-2=1$ , which agrees with the initial value S₁ = 1.

• Assume the closed form is correct for all n up to n = k. In particular, we assume

$S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}$

and

$S_k=3\times2^{k-1}+2(-1)^k$

We want to then use this assumption to show the closed form is correct for n = k + 1, or

$S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}$

From the given recurrence, we know

$S_{k+1}=S_k+2S_{k-1}$

so that

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)$

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}$

$S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)$

$S_{k+1}=3\times2^k-2(-1)^k$

$S_{k+1}=3\times2^k+2(-1)(-1)^k$

$\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}$

which is what we needed. QED

6 0

3 years ago

What is the approximate length of the diagonal of a square with side length of 20 centimeters

zhuklara [117]

Answer:28.28 cm

Explanation:

a² + b² = c²

20² + 20² = c²

c² = 400 + 400

c² = 800

c = √800

c = 28.28 cm

5 0

4 years ago

Read 2 more answers