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3 years ago

12

What is this? i’m so confused can you help me?

1 answer:

stealth61 [152]3 years ago

5 0

Answer:

can you give me the l.i.n.k to that?

Step-by-step explanation:

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Lulu ate 3/5 of a bag of chips which equaled to 150 calories. How much calories would have lulu ate if she ate the whole bag of

Lerok [7]

C=150•(3/5)•5
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So if she ate the whole bag, she’d eat 450 calories.

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3 years ago

WILL MARK BRAINLIEST ^^<br><br> picture included!! PLEASE and thank u

sineoko [7]

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3 years ago

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ludmilkaskok [199]

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3 years ago

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

schepotkina [342]

Answer:

<u>Question 11:</u>

$\angle DAC = 53^\circ$

$\angle AED = 90^\circ$

$\angle ADC = 74$

$DB = 16$

$AE = 6.03$

$AC = 12.06$

<u>Question 12:</u>

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

<u>Question 13: </u>

AC and BD are perpendicular lines, and they are diagonals

Step-by-step explanation:

<u>Question 11</u>

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

$AE = EC$

$AC = 6.03 +6.03$

$AC = 12.06$

<u>Question 12: Isosceles Triangle</u>

In the rhombus, all 4 sides are equal;

So, the isosceles triangle are:

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

<u>Question 13: </u>

AC and BD are perpendicular lines, and they are diagonals

4 0

3 years ago

If f(x) = x 2 + 1 and g(x) = 3x + 1, find 2f(1) + 3g(4). <br><br> 1. 45<br> 2. 43<br> 3. 42

nordsb [41]

<span>f(x) = x^2 + 1
f(1) = 1</span>^2 + 1 = 2
so 2f(1) = 2(2) = 4

<span>g(x) = 3x + 1
f(4) = 3(4) + 1 = 13
</span>3g(4) = 3(13) = 39

<span> 2f(1) + 3g(4) = 4 + 39 = 43
</span>
answer
<span>2. 43</span>

8 0

4 years ago