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2 years ago

An electrician has 4.1 meters of wire.

2 answers:

4 0

A)He can cut 5 strips of of 7/10m.

7/10=0.7m

0.7*5=3.5m
3.5m of wire wil be used in cutting possible 7/10m pieces.

b)Leftover will be 0.6m.
4.1-3.5=0.6m

topjm [15]2 years ago

3 0

We know that
$4.1m= \frac{41}{10}m$

Part a) <span>how many strip 7/10m long can he cut?
</span>
divide $\frac{41}{10}$ by $\frac{7}{10}$
so
$\frac{ \frac{41}{10}}{ \frac{7}{10}} = \frac{41}{7} \\ \\ =5.85 \\ \\ =5strip$

the answer part a) is
$5strip$

Part b) <span>how much wire will he have left over
</span> $5* \frac{7}{10} =3.5m$
$3.5$ m of wire wil be used in cutting possible $\frac{7}{10}$ m pieces

<span>so
wire l</span>eft over will be
$4.1-3.5=0.6 m$
<span>
the answer Part b) is
</span> $0.6m$

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3x + 7y = 21 solve for y

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• Move x to the other side

7y=21-3x

• Divide by 7

y = (21-3x)/7

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2 years ago

Leeza did this to solve an equation. Did she makes an error 4x=7+3(2x-5) 4x=7+6x-5

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Yes, there was a mistake in applying the distributive property. If you have to perform

$3(2x-5)$

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2 years ago

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Factor x^3 -7x^2 -5x+35 by grouping. what is the resulting expression

Anni [7]

Answer:

(x-7) (x^2-5)

Step-by-step explanation:

x^3 -7x^2 -5x+35

Make 2 groups

x^3 -7x^2 -5x+35

Factor x^2 from the first group and -5 from the second group

x^2 (x-7) -5(x-7)

Now factor (x-7) out

(x-7) (x^2-5)

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3 years ago

Which is not a factor of 48? A. 8 B. 14 C. 16 D. 24

alexira [117]

14 is not a factor of 48

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2 years ago

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Please dont ignore, Need help!!! Use the law of sines/cosines to find..

Ket [755]

Answer:

16. Angle C is approximately 13.0 degrees.

17. The length of segment BC is approximately 45.0.

18. Angle B is approximately 26.0 degrees.

15. The length of segment DF "e" is approximately 12.9.

Step-by-step explanation:

By the law of sine, the sine of interior angles of a triangle are proportional to the length of the side opposite to that angle.

For triangle ABC:

$\sin{A} = \sin{103\textdegree{}}$ ,
The opposite side of angle A $a = BC = 26$ ,
The angle C is to be found, and
The length of the side opposite to angle C $c = AB = 6$ .

$\displaystyle \frac{\sin{C}}{\sin{A}} = \frac{c}{a}$ .

$\displaystyle \sin{C} = \frac{c}{a}\cdot \sin{A} = \frac{6}{26}\times \sin{103\textdegree}$ .

$\displaystyle C = \sin^{-1}{(\sin{C}}) = \sin^{-1}{\left(\frac{c}{a}\cdot \sin{A}\right)} = \sin^{-1}{\left(\frac{6}{26}\times \sin{103\textdegree}}\right)} = 13.0\textdegree{}$ .

Note that the inverse sine function here $\sin^{-1}()$ is also known as arcsin.

By the law of cosine,

$c^{2} = a^{2} + b^{2} - 2\;a\cdot b\cdot \cos{C}$ ,

where

$a$ , $b$ , and $c$ are the lengths of sides of triangle ABC, and
$\cos{C}$ is the cosine of angle C.

For triangle ABC:

$b = 21$ ,
$c = 30$ ,
The length of $a$ (segment BC) is to be found, and
The cosine of angle A is $\cos{123\textdegree}$ .

Therefore, replace C in the equation with A, and the law of cosine will become:

$a^{2} = b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}$ .

$\displaystyle \begin{aligned}a &= \sqrt{b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}}\\&=\sqrt{21^{2} + 30^{2} - 2\times 21\times 30 \times \cos{123\textdegree}}\\&=45.0 \end{aligned}$ .

For triangle ABC:

$a = 14$ ,
$b = 9$ ,
$c = 6$ , and
Angle B is to be found.

Start by finding the cosine of angle B. Apply the law of cosine.

$b^{2} = a^{2} + c^{2} - 2\;a\cdot c\cdot \cos{B}$ .

$\displaystyle \cos{B} = \frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}$ .

$\displaystyle B = \cos^{-1}{\left(\frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}\right)} = \cos^{-1}{\left(\frac{14^{2} + 6^{2} - 9^{2}}{2\times 14\times 6}\right)} = 26.0\textdegree$ .

For triangle DEF:

The length of segment DF is to be found,
The length of segment EF is 9,
The sine of angle E is $\sin{64\textdegree}}$ , and
The sine of angle D is $\sin{39\textdegree}$ .

Apply the law of sine:

$\displaystyle \frac{DF}{EF} = \frac{\sin{E}}{\sin{D}}$

$\displaystyle DF = \frac{\sin{E}}{\sin{D}}\cdot EF = \frac{\sin{64\textdegree}}{39\textdegree} \times 9 = 12.9$ .

7 0

2 years ago