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3 years ago

The defeintion of equivalent ratios

Mathematics

1 answer:

Hitman42 [59]3 years ago

8 0

Tow ratios that have the same value when simplified.

Hope this helps!!;)

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A contractor purchases ceramic tile to remodel a kitchen floor. Each tile cost $4 and the adhesive/grouting material costs $17.8

erma4kov [3.2K]

Answer:

132 tiles

Step-by-step explanation:

Each tile cost $4

The adhesive/grouting material costs $17.82.

If the contractor pays a total of $545.82

Let the number of tiles be represented as x

The equation is represented as:

$17.82 + $4 × x = $545.82

17.82 + 4x = 545.82

Collect like terms

4x = 545.82 - 17.82

4x = 528

Divide bother sides by 4

4x/4 = 528/4

x = 132

The number of ceramic tiles the contractor bought = 132 tiles

8 0

3 years ago

Someone please help ASAP will give brainliest

algol13

√0.02 is rational

√2 is rational

√2 is irrational

√1/2 is rational

6 0

3 years ago

Need Help Answering this question.?!

sladkih [1.3K]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

If AB = 13, BC = 9, and CA = 17, list the angles in order from smallest to largest.

Licemer1 [7]

A,C,B
This is the answer;BC,AB,CA

8 0

4 years ago

Unsure how to do this calculus, the book isn't explaining it well. Thanks

krok68 [10]

One way to capture the domain of integration is with the set

$D = \left\{(x,y) \mid 0 \le x \le 1 \text{ and } -x \le y \le 0\right\}$

Then we can write the double integral as the iterated integral

$\displaystyle \iint_D \cos(y+x) \, dA = \int_0^1 \int_{-x}^0 \cos(y+x) \, dy \, dx$

Compute the integral with respect to $y$ .

$\displaystyle \int_{-x}^0 \cos(y+x) \, dy = \sin(y+x)\bigg|_{y=-x}^{y=0} = \sin(0+x) - \sin(-x+x) = \sin(x)$

Compute the remaining integral.

$\displaystyle \int_0^1 \sin(x) \, dx = -\cos(x) \bigg|_{x=0}^{x=1} = -\cos(1) + \cos(0) = \boxed{1 - \cos(1)}$

We could also swap the order of integration variables by writing

$D = \left\{(x,y) \mid -1 \le y \le 0 \text{ and } -y \le x \le 1\right\}$

and

$\displaystyle \iint_D \cos(y+x) \, dA = \int_{-1}^0 \int_{-y}^1 \cos(y+x) \, dx\, dy$

and this would have led to the same result.

$\displaystyle \int_{-y}^1 \cos(y+x) \, dx = \sin(y+x)\bigg|_{x=-y}^{x=1} = \sin(y+1) - \sin(y-y) = \sin(y+1)$

$\displaystyle \int_{-1}^0 \sin(y+1) \, dy = -\cos(y+1)\bigg|_{y=-1}^{y=0} = -\cos(0+1) + \cos(-1+1) = 1 - \cos(1)$

7 0

1 year ago