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3 years ago

What is the 8.356 rounded to the nearest hundredth

Mathematics

2 answers:

rodikova [14]3 years ago

5 0

Answer:

8.356 rounded to the nearest hundredth=8.360

KATRIN_1 [288]3 years ago

3 0

Answer:

8.36

Step-by-step explanation:

The hundredth decimal place is the second digit right of the decimal.

There should be 2 decimal places when rounded to the nearest hundredth.

If the third digit is less than 5, round down. If the third digit is 5 or more, round up.

In 8.356, the third digit is 6. 6 is 5 or more so we round the second digit up.

8.356 => 8.36

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The value of x in logx(9/63)=1/2 is

fenix001 [56]

Answer: x = 7/2

Step-by-step explanation: u juss simplify 1/2 x 7 to 7/2 so x = 7/2

8 0

4 years ago

PLEASE HELP!!!! 50 POINTS

Leni [432]

First you need to rewrite the equation in the form y = mx +b:

2y = x-4

Divide all terms by 2:

2y / 2 = x/2 - 4/2

Simplify:

y = 1/2x - 2

The slope of the line is 1/2.

A perpendicular line has a slope of the negative reciprocal, which would be -2.

Now using the point slope method:

y - y1 = m(x-x1)

Using the given point of (8,-4) and m = -2

you get:

y +4 = -2(x-8)

Simplify:

y +4 = -2x +16

Now subtract 4 from both sides to get the final equation:

y = -2x +12

3 0

3 years ago

Its awpicture please help if you can

ASHA 777 [7]

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4 0

3 years ago

RHOMBUS The diagonals of rhombus ABCD intersect at E. Given that

schepotkina [342]

Answer:

Question 11:

$\angle DAC = 53^\circ$

$\angle AED = 90^\circ$

$\angle ADC = 74$

$DB = 16$

$AE = 6.03$

$AC = 12.06$

Question 12:

$\triangle ABD$ , $\triangle BAC$ , $\triangle CDA$ and $\triangle DAB$

Question 13:

AC and BD are perpendicular lines, and they are diagonals

Step-by-step explanation:

Question 11

Given

$\angle BAC = 53^\circ$

$DE = 8$

See attachment for Rhombus

Required

Determine the indicated sides

Solving (a): $\angle DAC$

Diagonal CA divides $\angle DAB$ into 2 equal angles

i.e

$\angle DAC = \angle BAC$

So:

$\angle DAC = 53^\circ$

Solving (b): $\angle AED$

The angles at E is 90 degrees because diagonals AC and BD meet at a perpendicular.

So:

$\angle AED = 90^\circ$

Solving (c): $\angle ADC$

First, we calculate $\angle ADE$ , considering $\triangle ADE$ :

$\angle ADE + \angle AED + \angle DAC = 180$

$\angle ADE + 90 + 53 = 180$

$\angle ADE + 143 = 180$

$\angle ADE = -143 + 180$

$\angle ADE = 37$

To calculate $\angle ADC$ , we have:

$\angle ADC = 2*\angle ADE$

$\angle ADC = 2* 37$

$\angle ADC = 74$

Solving (d): $DB$

From the rhombus

$DB = DE +EB$

Where

$DE =EB$

So:

$DB = 8 + 8$

$DB = 16$

Solving (e): $AE$

To do this we consider $\triangle ADE$

Using the tan formula

$tan(\angle ADE) = \frac{AE}{DE}$

$\angle ADE = 37$ and $DE = 8$

So:

$\tan(37) = \frac{AE}{8}$

$AE = 8 * \tan(37)$

$AE = 6.03$

Solving (f): $AC$

This is calculated as:

$AC = AE + EC$

Where

$AE = EC$

$AC = 6.03 +6.03$

$AC = 12.06$

Question 12: Isosceles Triangle

In the rhombus, all 4 sides are equal;

So, the isosceles triangle are:

$\triangle ABD$ , $\triangle BAC$ , $\triangle CDA$ and $\triangle DAB$

Question 13:

AC and BD are perpendicular lines, and they are diagonals

4 0

3 years ago

Acompact disc box is 11.5 cm long,7 cm wide and 1.5 cm tall. What is the total surface area of a stack of five such compact disc

REY [17]

Answer:

1,172.5cm²

Step-by-step explanation:

Total surface area of a box = 2(LW + WH + LH)

L is the length = 11.5cm

W is the width = 7cm

H is the height = 1.5cm

Substitute

TSA = 2(11.5(7) + 7(1.5) + (11.5)(1.5))

TSA = 2(80.5+10.5+17.25)

TSA = 2(117.25)

TSA = 234.5cm²

Hence the total surface area of a stack of five such compact disc boxes will be 5(234.5) = 1,172.5cm²

5 0

3 years ago