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2 years ago

HELPPPP PLEASEEEEE

Mathematics

1 answer:

Alecsey [184]2 years ago

7 0

I’m not rly sure what it means by flipping the card …. I’m assuming there’s more to this question but if it’s what I think it is the only way this equation will be true is by switching the + Symbol to - (subtraction) which would be 1-2= 3-4 since it would 1=1 making the equation true

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Please help me with this ! ! ! ! !

melisa1 [442]

Answer:

First choice: m = -4

Step-by-step explanation:

The second equation is already solved for n, so lets's use the substitution method and substitute m - 2 for n in the first equation.

m - 2n = 8

m - 2(m - 2) = 8

m - 2m + 4 = 8

-m + 4 = 8

-m = 4

m = -4

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3 years ago

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An object starts at x=-15.8m and ends at x=20.3m. what was its displacement(unit=m)

liubo4ka [24]

36.1metres

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3 years ago

Which of the following graphs is the inverse of f(x) = x2 + 4?

Eddi Din [679]

Answer:

Step-by-step explanation:

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3 years ago

A new car that costs $17,500 loses 25 percent of its value in the first year. How much is the loss of the value.

EastWind [94]

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{25\% of 17500}}{\left( \cfrac{25}{100} \right)17500}\implies 4375$

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3 years ago

A heavy rope, 50 ft long, weighs 0.6 lb/ft and hangs over the edge of a building 120 ft high. Approximate the required work by a

Anastasy [175]

Answer:

Exercise (a)

The work done in pulling the rope to the top of the building is 750 lb·ft

Exercise (b)

The work done in pulling half the rope to the top of the building is 562.5 lb·ft

Step-by-step explanation:

Exercise (a)

The given parameters of the rope are;

The length of the rope = 50 ft.

The weight of the rope = 0.6 lb/ft.

The height of the building = 120 ft.

We have;

The work done in pulling a piece of the upper portion, ΔW₁ is given as follows;

ΔW₁ = 0.6Δx·x

The work done for the second half, ΔW₂, is given as follows;

ΔW₂ = 0.6Δx·x + 25×0.6 × 25 = 0.6Δx·x + 375

The total work done, W = W₁ + W₂ = 0.6Δx·x + 0.6Δx·x + 375

∴ We have;

W = $2 \times \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= 2 \times \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 750$

The work done in pulling the rope to the top of the building, W = 750 lb·ft

Exercise (b)

The work done in pulling half the rope is given by W₂ as follows;

$W_2 = \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 562.5$

The work done in pulling half the rope, W₂ = 562.5 lb·ft

6 0

2 years ago

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