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3 years ago

12

Which of the following comparisons is False

1 answer:

Verizon [17]3 years ago

4 0

A graph shows the first comparison is true, and the C temperature is cooler than the F temperature for the remaining comparisons.

The comparison that is FALSE is ...
C.....30 °C is warmer than 90 °F

_____
30 °C = 86 °F
32.2 °C ≈ 90 °F

You might be interested in

Find two consecutive positive integers such that the sum of their squares is 421 .

maria [59]

We assume the two numbers are x and (x+1)
so x^2+(x+1)^2=421
x^2+(x^2+2x+1)=421
2x^2+2x-420=0
According to the formula of quadratic
x=14 or-15
cuz we know the two numbers are integers
so x=14
therefore the other number is 15
To make sure that's correct
14^2+15^2=421

Hope that helps you!!

4 0

3 years ago

Given this expression:

Darya [45]

Answer:

the coefficient of x^2 is 16

Step-by-step explanation:

is the number in front of the variable and exponent

7 0

3 years ago

Given: PRST is a square

xxTIMURxx [149]

Answer:

$(1-\sqrt{2})a^2$

Step-by-step explanation:

Consider irght triangle PRS. By the Pythagorean theorem,

$PS^2=PR^2+RS^2\\ \\PS^2=a^2+a^2\\ \\PS^2=2a^2\\ \\PS=\sqrt{2}a$

Thus,

$MS=PS-PM=\sqrt{2}a-a=(\sqrt{2}-1)a$

Consider isosceles triangle MSC. In this triangle

$MS=MC=(\sqrt{2}-1)a.$

The area of this triangle is

$A_{MSC}=\dfrac{1}{2}MS\cdot MC=\dfrac{1}{2}\cdot (\sqrt{2}-1)a\cdot (\sqrt{2}-1)a=\dfrac{(\sqrt{2}-1)^2a^2}{2}=\dfrac{(3-2\sqrt{2})a^2}{2}$

Consider right triangle PTS. The area of this triangle is

$A_{PTS}=\dfrac{1}{2}PT\cdot TS=\dfrac{1}{2}a\cdot a=\dfrac{a^2}{2}$

The area of the quadrilateral PMCT is the difference in area of triangles PTS and MSC:

$A_{PMCT}=\dfrac{(3-2\sqrt{2})a^2}{2}-\dfrac{a^2}{2}=\dfrac{(2-2\sqrt{2})a^2}{2}=(1-\sqrt{2})a^2$

5 0

3 years ago

Emily lived in a blue house for 4.5 years. Then she lived in a yellow house for 3.6 years. How many years did Emily live in the

ikadub [295]

You add 3.6 plus 4.5 and you get 8.1 years

8 0

3 years ago

Read 2 more answers

Wilfred is feeding his goats. He gives 56% of the bag of feed to the goats in field A and the rest to the goats in field B. If t

krok68 [10]

Answer:

$Feeds = 11.76kg$

Step-by-step explanation:

Given

$Field\ A= 56\%$

$Field\ B= The\ rest$

$Bag = 21kg\ feed$

Required

Determine how much goats in field A, get.

To do this, we simply multiply the percentage feed received by goats in field A by the actual bags of feeds

i.e.

$Feeds = 56\% * 21kg$

Convert percentage to decimal

$Feeds = 0.56 * 21kg$

$Feeds = 11.76kg$

<em>Hence, goats in field A get 11.76kg of feeds</em>

6 0

2 years ago