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3 years ago

What is the value of 7m - 3n when m = 3 and n = 2? A. 5 B. 15 C. 24 D. 36

Mathematics

2 answers:

Bezzdna [24]3 years ago

5 0

Well just plug those two numbers into the equation: 7(3) - 3(2) Which equals 15 so B) 15

lawyer [7]3 years ago

5 0

7(3)-3(2)
21-6= 15
The answer would be B.

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there are 27 red or blue marbles in the bag the number of red marbles is five less than three times the number of the blue marbl

alexdok [17]

So assuming that the total =27
r+b=27
r=-5+3b
r=3b-5
subsitute 3b-5 for r in first equation
3b-5+b=27
4b-5=27
add 5
4b=32
divide by 4
b=8
subsitute
b+r=27
8+r=27
subtracct 8
r=19

red=19
blue=8

4 0

3 years ago

a submarine is moving at a 35 km/h. It can say underwater for another 4.5 hours. If it speeds a steady, how long will it take a

dedylja [7]

Answer:

1 hour 24 minutes

Step-by-step explanation:

49/35=1.4

8 0

4 years ago

What is the volume of a hemisphere with a radius of 26.7 m, rounded to the nearest tenth of a cubic meter?

attashe74 [19]

Answer:

$39,865.1m^3$

Step-by-step explanation:

a hemisphere is half a sphere.

And if the formula for the volume of a sphere is:

$V=\frac{4\pi r^3}{3}$

we must calculate the volume of the sphere and in the end divide by two to find the volume of the hemisphere.

Since the radius is:

$r=26.7m$

substituting this into the formula for volume:

$V=\frac{4\pi (26.7m)^3}{3} \\\\V=\frac{4(3.1416)(19,034.163m^3)}{3} \\\\V=79,730.12m^3$

this is the volume of the total sphere, thus the volume of a hemisphere is:

$\frac{V}{2}=\frac{79,730.12m^3}{2}\\ =39,865.06m^3$

rounding to the nearest tenth:

$39,865.1m^3$

8 0

4 years ago

Read 2 more answers

Find the first three iterates of the function f(z) = z2 + c with a value of c = 2 - 3i and an initial value of z0 = 1 + 2i.

Artist 52 [7]

Answer:

Step-by-step explanation:

We have: (I rewrote the function)

$f(z_n)={z_{n-1}} ^2+c$

Given that:

$\displaystyle c=2-3i \text{ and } z_0 = 1 + 2 i$

The first iterate will be:

$\displaystyle \begin{aligned} f(z_1)&=(z_0)^2+c \\ &=(1+2i)^2+(2-3i) \\ &= (1+4i+4i^2)+(2-3i) \\ &=1+4i-4+2-3i \\ &=-1+i \end{aligned}$

The second iterate will be:

$\begin{aligned}f(z_2) &=(z_1)^2+c\\ &=(-1+i)^2+(2-3i) \\&= (1-2i+i^2)+(2-3i) \\&=1-2i-1+2-3i \\&=2-5i \end{aligned}$

And the third iterate will be:

$\begin{aligned} f(z_3)&=(z_2)^2+c\\ &=(2-5i)^2+(2-3i) \\ &=(4-20i+25i^2)+(2-3i) \\ &=4-20i-25+2-3i \\ &=-19-23i \end{aligned}$

Hence, our answer is C.

3 0

3 years ago

Factorize 2a^2+7a-15

loris [4]

$\huge\bf \pink {\underline{Solution :-}}$

$2 {a}^{2} + 7a - 15$

$= 2 {a}^{2} + 10a - 3a - 15$

$= 2a(a + 5) - 3(a + 5)$

$= (a + 5)(2a - 3)$

$\sf \red{Hence, Answer \: is \: (a + 5)(2a - 3).}$

$\\$

$\bf \purple{ \underline{ Important \: Formulas \: for \: Factorization :-}}$

$• \: {a}^{2} + 2ab + {b}^{2} = {(a + b)}^{2}$

$• \: (a + b)(a - b) = {a}^{2} - {b}^{2}$

$• \: {a}^{2} - 2 ab + {b}^{2} = {(a - b)}^{2}$

$• \: {a}^{3} + 3 {a}^{2} b + 3a {b}^{2} + {b}^{3} = {(a + b)}^{3}$

$• \: {a}^{3} - 3 {a}^{2} b + 3a {b}^{2} - {b}^{3} = {(a - b)}^{3}$

$• \: {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} )$

$• \: {(a + b)}^{2} - {(a - b)}^{2} = 4ab$

$• \: (a + b)( {a}^{2} -ab + {b}^{2} ) = {a}^{3} + {b}^{3}$

$• \: (a - b)( {a}^{2} + ab + {b}^{2} ) = {a}^{3} - {b}^{3}$

$• \: {( \frac{a + b}{2} )}^{2} - ( {\frac{a - b}{2} } )^{2} = ab$

5 0

3 years ago

Read 2 more answers