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Cloud [144]
3 years ago
10

Solve the equation:  4 1/5 + b = 9 3/5

Mathematics
2 answers:
Yanka [14]3 years ago
7 0
4 \frac{1}{5}+ b = 9 \frac{3}{5} \ \ | (-4\frac{1}{5})\\ \\4 \frac{1}{5 }-4 \frac{1}{5}+ b = 9 \frac{3}{5}-4 \frac{1}{5}\\ \\b= 5\frac{2}{5}=5.4


kvasek [131]3 years ago
3 0
4 \frac{1}{5}  + b = 9  \frac{3}{5} \\\\b= 9  \frac{3}{5}-4 \frac{1}{5} \\\\b=5 \frac{2}{5}
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likoan [24]

Answer:

4c - 2

Step-by-step explanation:

Add up all the terms to find the perimeter.

2c + 2(c - 1)

2c + 2c - 2

4c - 2

Therefore, the perimeter is 4c - 2.

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3 years ago
The area of a particular rectangle is 72. If the length of the rectangle is twice
soldier1979 [14.2K]

Answer:

6

Step-by-step explanation:

Suppose the width is "a".  Then the length is 2a.

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4 0
3 years ago
What is the answer please.... need help
mina [271]

Answer:

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

f\left(x\right)=x^3-6x^2+3x+10

As the highest power of the x-variable is 3 with the leading coefficients of 1.

  • So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

f\left(x\right)=x^3-6x^2+3x+10

0=x^3-6x^2+3x+10              ∵  f(x)=0

as

Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0

so

\left(x+1\right)\left(x-2\right)\left(x-5\right)=0    

Using the zero factor principle

if  ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0

x=-1,\:x=2,\:x=5

Therefore, the zeros of the function are:

x=-1,\:x=2,\:x=5

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.    

8 0
3 years ago
xand y are light house. y being 20km due east of x and from a ship due south of x the bearing of y was 055°. what is the distanc
leonid [27]

Answer:

I) |xz| ≈ 28.6 km

II) |yz| ≈ 34.8 km

Step-by-step explanation:

Let's assume that the position of ship due south of x is z (aà pictor representation of the question is attached)

|xy| = 20 km, |xz| = ?, |yz| = ?, θ(y) = 55°

Using Trigonometric ratio - SOHCAHTOA

I) Tan θ = |xz| ÷ |xy| ⇒ Tan 55° = |xz| ÷ 20

|xz| = 20 * Tan 55 = 20 * 1.428

|xz| = 28.56 km

|xz| ≈ <u>28.6 km</u>

<u />

II) Cos θ = |xy| ÷ |yz| ⇒ Cos 55° = 20 ÷ |yz|

|yz| * Cos 55° = 20 ⇒ |yz| = 20 ÷ Cos 55°

|yz| = 20 ÷ 0.574 = 34.84 km

|yz| ≈ <u>34.8 km</u>

4 0
3 years ago
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