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2 years ago

Find the perimeter and area of the figure. Round to the nearest tenth if necessary.

Mathematics

1 answer:

Stella [2.4K]2 years ago

3 0

Answer: D

Step-by-step explanation:

$A=\frac{1}{2}bh\\\\A=\frac{1}{2}(4.8)(7.2)\\\\A \approx \boxed{17.3}$

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Ted needs an average of at least 70 on his three history test he has scored already 85 and 60 on two test what is the minimum gr

PilotLPTM [1.2K]

So average=(total of scores)/(number of tests)
needs at least average of 70
at least is represented as greater than or equal to or the sign (>)
70>(total)/(number oftests)

since we have 3 tests, we have to have 3 scores so
70>(x+y+z)/3
he scored 85 and 60
70>(x+85+60)/3 (doesn't matter which to subsitute)
70>(x+145)/3
multiply obht sides by 3
210>x+145
subtract 145 from both sides
65>x
he needs to get at leas 65 on his third test

4 0

3 years ago

80 cm to 42.5 cm for percent change

Fofino [41]

Answer:

= 46.875% decrease

Step-by-step explanation:

Mark as brainliest :)

5 0

3 years ago

Plzz help I will mark brainliesttttt

ira [324]

Answer:

HOPE IT HELPED U .....

Step-by-step explanation:

a). f(5)= 2x +6

= 2 *5+6

= 10+6

=16

(b). g (9)= -12x+4

= -12*9+4

= -108 +4

=-104

(c). f(5)+f(9)= 16+(-104)

=-88

4 0

3 years ago

Read 2 more answers

Drag each number to the box on the right so it matches the value of the expression on the left. Each number may be used once, mo

Ymorist [56]

Answer:

$-62024.48$

Step-by-step explanation:

Given

$[ 4 * ( 4 + 3 ) ] - 3 [ ( 7 - 5 ) \div 1 2 ] * 2 ( 5.4 + 3 ) * ( 6 - 2 ) [ 9 \div 3 + 6 ] * 3 ( 4 + 3.2 ) * [ ( 9 - 2 ) + 2.5]$

Required

Solve

The question is poorly formatted; the only way to answer this is to place the result of each step in brackets until brackets cannot be used.

So, we have

$[ 4 * ( 4 + 3 ) ] - 3 [ ( 7 - 5 ) \div 1 2 ] * 2 ( 5.4 + 3 ) * ( 6 - 2 ) [ 9 \div 3 + 6 ] * 3 ( 4 + 3.2 ) * [ ( 9 - 2 ) + 2.5 ]$

Solve the expressions in the inner brackets

$[ 4 * 7 ] - 3 [ 2 \div 1 2 ] * 2 ( 8.4 ) * ( 4 ) [ 9 \div 3 + 6 ] * 3 ( 7.2 ) * [ 7 + 2.5 ]$

Solve all divisions

$[ 4 * 7 ] - 3 * \frac{1}{6} * 2 ( 8.4 ) * ( 4 ) [ 3 + 6 ] * 3 ( 7.2 ) * [ 7 + 2.5 ]$

$[28] - 0.5 * (16.8) * (4) [9] * (21.6) * [9.5]$

$[28] - 0.5 * (16.8) * [36] * (21.6) * [9.5]$

Multiply through

$28- 62052.48$

$-62024.48$

3 0

3 years ago

A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,

zloy xaker [14]

Answer:

$t=-20ln\left(\dfrac{1}{3}\right)$

Step-by-step explanation:

The relationship between A, the area of the glacier in square kilometers, and t, the number of years the glacier has been melting, is modeled by the equation.:

$A=45e^{-0.05t}$

We want to determine the value of t for which the area, A(t)=15 square kilometers.

$15=45e^{-0.05t}\\$Divide both sides by 45\\\dfrac{15}{45} =\dfrac{45e^{-0.05t}}{45}\\\dfrac{1}{3}=e^{-0.05t}\\$Take the natural logarithm of both sides\\ln\left(\dfrac{1}{3}\right)=ln\left(e^{-0.05t}\right)\\ln\left(\dfrac{1}{3}\right)=-0.05t\\$Divide both sides by -0.05$\\t=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05} \\=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05}\\t=-20ln\left(\dfrac{1}{3}\right)$

Therefore, the time for which the area will be 15 sqyare kilometers is:

-20 ln(1/3) years.

7 0

3 years ago