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4 years ago

Students must mix blue and yellow paint to make different shades of green. The shade of green

Mathematics

1 answer:

denis-greek [22]4 years ago

6 0

This non-question cant be answered without knowing what the table looks like

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Four balls of wool will make 8 knitted caps. How many balls of wool will Malcolm need if he wants to make 6 caps

netineya [11]

He would only need 3 because
Knitted Caps=2
so that means
4(2) Knitted caps=8
3(2) knitted caps = 6

7 0

3 years ago

Solve for x. NEED ANSWER ASAP TYY SO MUCH IN ADVANCE

Serjik [45]

Answer:

x = 10√3

Step-by-step explanation:

Because of the right angles, we can use Pythagoras

30² - x² = y² (a) Largest triangle

y² - 20² = h² (b) Medium triangle

substitute y² from (a) into (b)

30² - x² - 20² = h²

500 - x² = h² (d)

x² - h² = (30 - 20)² (c) smallest triangle

substitute h² from (d) into (c)

x² - (500 - x²) = 100

2x² = 600

x² = 300

x = 10√3

4 0

3 years ago

If bob earns 35,000 a month how much does he earn in 2 years and 75 weeks

nlexa [21]

Okay, do 35,000x24=840,000
840,000/104=8,077
8077x75=605,769
605,769+840,000=1,445,769.23
:-)

6 0

3 years ago

Read 2 more answers

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago

What is the least common multiple of 8 and 18? A 16 B 36 С 72 D 144

Verdich [7]

Answer:72

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers