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3 years ago

Standard form for 78 hundreds 58 tens 19 ones

Mathematics

1 answer:

Nastasia [14]3 years ago

4 0

Answer:

8399

Step-by-step explanation:

78*100=7800

58*10=580

7800+580+19=8399

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1. take away five from twelve times f. 2. one-half of the sum of k and six 3.x squared minus the sum of 5 4.the sum of the produ

77julia77 [94]

The expressions formed are,

12f -5
(k+6)/2
x²-x+5
ab+3c
24xy+g

Formation of expressions in 1, 2 and 3:

In 1, twelve times f is, 12f

Taking away 5, it becomes (12f-5)

In 2, sum of k and 6 is, (k+6)

One-half of the above quantity is, (k+6)/2

In 3, sum of 5 with x is, (x+5)

Now, x squared minus the above expression indicates (x²-x+5)

Formation of expressions in 4 and 5:

In 4, product of a and b, is ab and 3 times c is 3c

Sum of the expressions evaluated in the previous statement = ab+3c

In 5, 24 times the product of x and y is, 24xy

Adding, g in the above computed expression, we get, 24xy+g

Learn more about expressions here:

brainly.com/question/17167810

#SPJ4

3 0

1 year ago

2 tan 30°<br>II<br>1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago

I NEED THE ANSWER ASAP PLEASE

Natali [406]

Answer:

720$

Step-by-step explanation:

(1000+(230*24))-5800=720

8 0

2 years ago

A rectangular pool 12 meters by 8 meters is surrounded by a walkway of width x meters. At what

Thepotemich [5.8K]

X=3

x is the width of the sidewalk
The outside dimensions of the walkway are 12 + 2x by 8 + 2x
The total area is (12 + 2x)(8 + 2x)
The total area minus the area of the pool should be equal to the area of the pool

Hoped this helped!

8 0

2 years ago

A table showing Pounds, Price per Pound, and Total Cost. The first row shows Cranberries, with the entries 1, 4, and 1 times 4.

Alex Ar [27]

Answer:

4.80

Step-by-step explanation:

Just Did It.

3 0

2 years ago

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