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ArbitrLikvidat [17]

2 years ago

14

Which of the following equations represents the graph shown?

2 answers:

zhenek [66]2 years ago

8 0

He’s correct .yay ya

Sonja [21]2 years ago

5 0

Answer:

It's the first one. -x+2

Step-by-step explanation:

The slope is negative and the line intersects the y-intercept at positive 2.

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Factor by Grouping 8x+8y+rx+ry<br> I do not understand this.

nadezda [96]

FACTORISATION IS BASICALLY FINDING LIKE-TERMS.
1. FIND THE HCF OF 8. HERE, THE HCF IS 8. ALSO, CHECK ALL THE LIKE-TERMS. LIKE-TERMS ARE OUTSIDE THE BRACKET AND UNLIKE TERMS ARE INSIDE.
2. ADD THE TERMS IN THE BRACKET AND THE TERMS OUTSIDE THE BRACKET FROM STEP 1.

8x + 8y + rx + ry
1. 8 (x + y) + r (x + y)
2. (x +y) (8+r)

8 0

2 years ago

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The sum of two numbers is 225. The difference of the same two numbers is 35.What are the 2 numbers?

Ksivusya [100]

Answer:

130 and 95

Step-by-step explanation:

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4 0

2 years ago

If these two figures are similar, what is the measure of the missing angle? 180 90 70 110

yawa3891 [41]

Answer:

70

Step-by-step explanation:

8 0

3 years ago

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There is 28 kg 750 g of sugar in a sack. How many tins of 500 g sugar can be filled using the sugar in the sack?

vovangra [49]

Answer:

Step-by-step explanation:

28 kg + 750 g = 28*1000 + 750

28000 + 750

= 28750 g

Number of sack = Total quantity of sugar ÷ quantity of sugar in a sack

= 28750 ÷ 500

= 57 sacks

6 0

2 years ago

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Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant

pychu [463]

Answer:

A=152

K= -Ln(0.5)/14

Step-by-step explanation:

You can obtain two equations with the given information:

T(14 minutes) = 114◦C

T(28 minutes)=152◦C

Therefore, you have to replace t=14, T=114 and t=28, T=152 in the given equation:

$114=190-Ae^{-14k} (I) \\152=190-Ae^{-28k}(II)$

Applying the following property of exponentials numbers in (II):

$e^{a}.e^{b}=e^{a+b}$

Therefore $e^{-28k}$ can be written as $e^{-14k}.e^{-14k}$

$152=190-Ae^{-14k}.e^{14k}$

Replacing (I) in the previous equation:

$152=190-76e^{-14k}$

Solving for k:

Subtracting 190 both sides, dividing by -76:

$0.5=e^{-14k}$

Applying the base e logarithm both sides:

Ln(0.5)= -14k

Dividing by -14:

k= -Ln(0.5)/14

Replacing k in (I) and solving for A:

$Ae^{-14(-Ln(0.5)/14)}=76\\Ae^{Ln(0.5)} =76\\A(0.5)=76$

Dividing by 0.5

A=152

7 0

2 years ago