Mr. Nasser asks four of his students to measure four different objects. Here are the results.

Pam likes to practice dancing while preparing for a math tournament. She spends 80 minutes every day practicing dance and math.

SVETLANKA909090 [29]

Answer:

Part A)

x+y=80

y=x+20

B)

30 minutes

C)

no

If Pam was to spend 60 minutes practicing dance, then she would only get 20 minutes of math. 60-20=40. She would be practicing dance 40 minutes more than math, not just 20.

3 0

2 years ago

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A rectangular floor tile is shown it dimensions are given to the nearest 0.1 metres the tile is only able to sustain maximum pre

egoroff_w [7]

Answer:

736 N

Step-by-step explanation:

The dimensions of the rectangular tile are:

Length = 2.3m

Width = 1.6m

The pressure exerted on a surface is given by the formula

$p=\frac{F}{A}$

where

p is the pressure

F is the force exerted

A is the area on which the force is exerted

In this problem, we have:

$p=200 N/m^2$ is the maximum pressure that the tile is able to sustain

A is the area of the tile, which can be calculated as the product between length and width, so:

$A=L\cdot w =(2.3)(1.6)=3.68 m^2$

Re-arranging the formula for F, we can find the maximum force that can be safely applied to the tile:

$F=pA=(200)(3.68)=736 N$

4 0

2 years ago

The volumes of two similar figures are 343 mm3 and 512 mm3. If the surface area of the larger figure is 192 mm2, what is the sur

Kobotan [32]

In geometry, similar figures are those whose ratios of the corresponding sides are equal and the corresponding angles are congruent. In relation to the volume, we determine first the cube roots of the given and find the ratio as shown below.

s1 / s2 = cube root of (512/343)
= 8/7
The square of this ratio is the ratio of the areas of the figure. If we let x be the area of the smaller figure then,
(8/7)^2 = 192 mm²/ x
The value of x from the equation is 147 mm².

The area therefore of the smaller figure is 147 mm².

3 0

2 years ago

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A force of 500.0 is represented graphically with its tail at the origin and the tip pointed in a direction 30.0° above the posit

trapecia [35]

Answer:

$F^{'}=(250\sqrt{3},250 })$