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stealth61 [152]
3 years ago
12

Plz help what do i do and how do i do this its due tomorrow

Mathematics
1 answer:
Aleks [24]3 years ago
6 0
In order to find the area, you multiply the width (7), height (6) and length (12). But the problem states that the height has changed from 6 to 6 minus3, which would be 3. So you multiply 3, 7, and 12.
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46 divided by 5 long divison
DIA [1.3K]
9.2 (b) is the answer hope this helped!
5 0
3 years ago
Tom begins his morning shift at 9:20 A.M. He works for 3 hours. He then takes a 50 minute lunch break before working his afterno
lys-0071 [83]
9:20 + 3:00= 12:00
12:00 + 0:50= 12:50 A.M. Tom's afternoon shift begins
7 0
4 years ago
SEVENTEEN POINTS IF ANSWERED CORRECTLY!!
grandymaker [24]
As BD is perpendicular....given

ADB AND CDB are 90° ......as BD is perpendicular so 90°

BD= BD .....common line !

4 0
3 years ago
Read 2 more answers
3. Marcus's family has completed 30% of a trip. They have traveled 21 miles.
vlada-n [284]

Answer: 70 miles

Step-by-step explanation: scince the trip is 30% finished and that 30% was 21 miles you just have to multiply 21 by 3 which is 63and than you have to figure out what 1 third of 21 is which is 7 and than you just have to add 7 to 63. Hope this helps

3 0
3 years ago
What are expressions for MN and LN? Hint Construct the altitude from M to LN.
Nikitich [7]

The question is missing the figure. So, it is in the atachment.

Answer: MN = x\sqrt{2}  LN = \frac{x}{2}.(\sqrt{2} + \sqrt{6} )

Step-by-step explanation: The first figure in the attachment is the figure of the question. The second figure is a way to respond this question by tracing the altitude from M to LN as suggested. When an altitude is drawn, it forms a 90° angle with the base, as shown in the drawing. To determine the other angle, you have to remember that all internal angles of a triangle sums up to 180°.

For the triangle <u>on the left</u> of the altitude:

45+90+angle=180

angle = 45

For the triangle <u>on the right</u>:

30+90+angle=180

angle = 60

With the angles, use the Law of Sines, which is relates sides and angles, as follows:

\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}

For MN:

\frac{x}{sin(30)} = \frac{MN}{sin(45)}

MN = \frac{x.sen(45)}{sen(30)}

MN = x\sqrt{2}

For LN:

\frac{LN}{sen(105)} =\frac{x}{sin(30)}

LN = \frac{x.sin(105)}{sin(30)}

We can determine sin (105) as:

sin(105) = sin(45+60)

sin(105) = sin(45)cos(60) + cos(45)sin(60)

sin(105) = \frac{\sqrt{2} }{2}.\frac{1}{2} + \frac{\sqrt{2} }{2}.\frac{\sqrt{3} }{2}

sin(105) = \frac{\sqrt{2} }{4} + \frac{\sqrt{6} }{4}

LN = \frac{x.sin(105)}{sin(30)}

LN = x.(\frac{\sqrt{2} }{4} + \frac{\sqrt{6} }{4}  ) .2

LN = \frac{x}{2}.(\sqrt{2} + \sqrt{6} )

The expressions for:

MN = x\sqrt{2}

LN = \frac{x}{2}.(\sqrt{2} + \sqrt{6} )

6 0
3 years ago
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