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2 years ago

Solve in terms of d. 6dh = 2y

Mathematics

1 answer:

sukhopar [10]2 years ago

7 0

Answer:

d = $\frac{2y}{6h}$

Step-by-step explanation:

Given

6dh = 2y ( isolate d by dividing both sides by 6h )

d = $\frac{2y}{6h}$

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Jerry goes to a theme park to ride the roller coasters. The theme park charges an entry fee in addition to a fee for each roller

ratelena [41]

You start with: (assuming x equals the cost to enter and y the cost of going on the rollercoasters.)
x+5y=35
x+11y=59. Multiply the top equation by -1, and subtract the equations, giving you -6y=-24, divide by -6 into both sides of the equation, to get y=4. Now replace y in one of the original equations (I recommend x+5y=35) and solve for x, giving you x=15

The cost for entering is 15 dollars, while each coaster is 4 dollars more. You could simplify this by changing y into x and making it slope-intercept form, to track your cost. y=4x+15, so it has a slope of 4, and a y-intercept of 15. This answer should give you a good grade on a test.

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2 years ago

EASY POINTS EVALUATION

dedylja [7]

Answer:

$\frac{4}{9}$

Step-by-step explanation:

$( \frac{16 {}^{ \frac{1}{2} } }{81 { }^{ \frac{1}{2} } }) \\ ( \frac{ {4}^{2 \times \frac{1}{2} } }{9^{2 \times \frac{1}{2} } }) \\ ( \frac{4^{1} }{9 ^{1} } ) \\ \frac{4}{9}$

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1 year ago

Read 2 more answers

"assume that s is a non-empty set of real numbers which is bounded above and lambda is the least upper bound. Prove that for all

s2008m [1.1K]

By definition, if $\lambda$ is the least upper bound of the set $S$ , it means two thing:

$\forall x \in S,\ x \leq \lambda$
$\forall \varepsilon>0,\ \exists x \in S:\ x>\lambda-\varepsilon$

In other words, the least upper bound of a set is greater than or equal to every single element of the set, but it is "close enough" to the elements of the set, because you guaranteed to find elements in the set between $\lambda-\varepsilon$ and $\lambda$

For example, pick $S = [1,10)$ . Obvisouly, the least upper bound is $\lambda = 10$ . In fact, every number in $[1,10)$ is smaller than 10, but as soon as you take away something from 10, say 0.01, you get 9.99, and there are elements in $S$ greater than 9.99, say 9.9999.

So, the claim is basically proven by definition: if $\beta < \lambda$ , let $0 < \delta = \lambda - \beta$ . By definition, there exists $\alpha \in S:\ \alpha > \lambda - \delta$ .

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3 years ago

Oliver earns $23.80 an hour. He earns normal rates during week days and time-and-a-

Oxana [17]

So first we calculate the weekly hours:
23.80 x 34 = 809.20

Then we calculate the time and half:
23.80 x 6 x 1.5 = 214.2

Then we total then together:
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2 years ago

A refrigerator costs $429 to buy and its label says that estimated energy costs are $43 per year. Another brand costs $399, but

Nezavi [6.7K]

Answer:

(6, 687)

Step-by-step explanation:

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3 years ago