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3 years ago

Explain why it can be helpful to use partial quotients when dividing?

1 answer:

3 0

In mathematics, a quotient (from Latin: quotiens "how many times", pronounced ˈkwoʊʃənt) is the result of division. For example, when dividing 6 by 3, the quotient is 2, while 6 is called the dividend, and 3 the divisor.

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30% of what number is 10.2

rewona [7]

30.6 because 10.2 times 30 is 30.6

7 0

4 years ago

Read 2 more answers

Given g(x)= 5/x+3, find g(-3).

kherson [118]

5/0

If you put in -3 for x then -3+3 is 0. You can't divide by zero so it's no solution if you have that option!

5 0

4 years ago

Cual es el resultado correcto de esta expresión aritmética? 18-12+2×3² - 12+(1+1)=?

melisa1 [442]

Answer:

3,6,9,12,15,18,21}

B={4,8,12,16,20}

C={2,4,6,8,10,12,14,16}

D={5,10,15,20}

(i) A−B={3,6,9,15,18,21}

(ii) A−C={3,9,15,18,21}

(iii) A−D={3,6,9,12,18,21}

(iv) B−A={4,8,16,20}

(v) C−A={2,4,8,10,14,16}

(vi) D−A={5,10,20}

(vii) B−C={20}

(viii) B−D={4,8,12,16}

(ix) C−B={2,6,10,14}

(x) D−B={5,10,15}

(xi) C−D={2,4,6,8,12,14,16}

8 0

3 years ago

If a and b are two angles in standard position in Quadrant I, find cos(a+b) for the given function values. sin a=15/17and cos b=

tensa zangetsu [6.8K]

The value of $cos(a+b)$ for the angles $a$ and $b$ in standard position in the first quadrant is $-\frac{36}{85}$

We need to find the value of $cos(a+b)$ . To proceed, we need to use the compound angle formula

<h3>Cosine of a sum of two angles</h3>

The cosine of the sum of two angles $a$ and $b$ is given below

$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$

We are given

$sin(a)=\dfrac{15}{17}\\\\cos(b)=\dfrac{3}{5}$

We need to find $sin(b)$ and $cos(a)$ , using the identity

$sin^2(\theta)+cos^2(\theta)=1$

To find $sin(b)$ , note that

$sin^2(b)+cos^2(b)=1\\\\\implies sin(b)=\sqrt{1-cos^2(b)}$

substituting $\frac{3}{5}$ for $cos(b)$ in the identity, we get

$sin(b)=\sqrt{1-cos^2(b)}\\\\=\sqrt{1-\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}$

To find $cos(a)$ , note that

$sin^2(a)+cos^2(a)=1\\\\\implies cos(a)=\sqrt{1-sin^2(a)}$

substituting $\frac{15}{17}$ for $sin(a)$ in the identity, we get

$cos(a)=\sqrt{1-sin^2(a)}\\\\=\sqrt{1-\left(\dfrac{15}{17}\right)^2}=\dfrac{8}{17}$

<h3>Find the value of cos(a+b)</h3>

We can now make use of the formula

$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$

to find $cos(a+b)$ .

$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)\\\\=\dfrac{8}{17}\cdot\dfrac{3}{5}-\dfrac{15}{17}\cdot\dfrac{4}{5}=-\dfrac{36}{85}$

Learn more about sine and cosine of compound angles here brainly.com/question/24305408

8 0

2 years ago

The formula for the area of a trapezoid is A = one-half (b Subscript 1 Baseline + b Subscript 2 Baseline) times h When this equa

Andre45 [30]

Answer:

The answer is \frac{2A}{h} -b_2=b_1

Step-by-step explanation:

Given that the area of a trapezoid is ;

A=1/2 (b₁+b₂)h

When the equation is solved for b₁ it will be;

A=1/2 (b₁+b₂)h

This can be written as;

A=h/2 (b₁+b₂)

Multiply both sides by 2/h

2A/h = b₁+b₂

Make b₁ subject of the formula

2A/h - b₂ = b₁

3 0

3 years ago