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3 years ago

1. At the end of one school day a teacher had 17 crayons left. The teacher remembered

Mathematics

2 answers:

Vlad [161]3 years ago

5 0

Answer:

30 crayons

Step-by-step explanation:

Let x be the number of crayons he started with

gave out 14 crayons

x-14

Got 12 back

x-14+12

Gave out 11 after lunch

x-14+12 -11

This equals 17

x-14+12 -11 =17

Combine like terms

x-13 = 17

Add 13 to each side

x -13+13 =17+13

x = 30

Andru [333]3 years ago

5 0

Answer: 30

Step-by-step explanation:

For this problem work backwards. Start from 17 and add 14. You should get 31. Then subtract 12, which equals 19. Finally add 11 to 19, which equals 30. Basically you are doing the inverse operation to get your answer. Hope this helps!

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Find the volume of the solid. 6 cm 8cm 15 cm

Bess [88]

Answer:

720 cm^3

Step-by-step explanation:

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3 years ago

Find the quotient of 5+4i/6+8i , and express it in the simplest form

Sergio [31]

Remember: i² = -1

$\frac{5+4i}{6+8i}= \frac{(5+4i)(6-8i)}{(6+8i)(6-8i)}= \frac{30-40i+24i-32i^2}{36-64i^2}= \frac{30-16i-32(-1)}{36-64(-1)}= \\ \\ = \frac{30-16i+32}{36+64}= \frac{62-16i}{100}=0.62-0.16i$

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7 0

3 years ago

Help me on this pleasseeee

EastWind [94]

Answer:

1) E and F

Step-by-step explanation:

So width is x

legth Is 3x+3

Perimeter is

3x+3+3x+3+x+x

Now simplify

8x+6

Now go through each choice that equals 8x+6

Only e and f

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2 years ago

Read 2 more answers

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Dennis_Churaev [7]

1. Is C 2. Is A 3. Might be D 4. Is B 5. Is D

8 0

2 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.

3 0

3 years ago