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2 years ago

9

Jessica spent S28 on fruit at the grocery store. She spent a total of $40 at the store. What percentage of the total did she spe

nd on fruit?

1 answer:

Gennadij [26K]2 years ago

6 0

Answer:

Jessica spent S28 on fruit at the grocery store. She spent a total of $40 at the store. What percentage of the total did she spend on fruit?

Step-by-step explanation:

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What is the probability that If I toss a die, I land on the number 6, three<br> times in a row?

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Answer:

Your Answer would 1/6

Step-by-step explanation:

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3 years ago

The sum of two binomials is 3z-6. If one binomial is z+4, what is the other binomial?

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The answer is 2z-10. Hope it helps.

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Joyce paid 180$ for an item at the store that was 20 percent of the original price. What was the original price

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3 years ago

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Given: KLMN is a trapezoid, m∠N=m∠KML, FD=8, LM KN = 3 5 F∈ KL , D∈ MN , ME ⊥ KN KF=FL, MD=DN, ME=3 5 Find: KM

denis23 [38]

Answer:

The length of side KM is $\sqrt{109}$ units.

Step-by-step explanation:

Given information: KLMN is a trapezoid, ∠N= ∠KML, FD=8, LM:KN=3:5, F∈ KL, D∈ MN , ME ⊥ KN KF=FL, MD=DN, $ME=3\sqrt{5}$ .

From the given information it is noticed that the point F and D are midpoints of KL and MN respectively. The height of the trapezoid is $3\sqrt{5}$ .

Midsegment is a line segment which connects the midpoints of not parallel sides. The length of midsegment of average of parallel lines.

Since LM:KN=3:5, therefore LM is 3x and KN is 5x.

$\frac{3x+5x}{2}=8$

$\frac{8x}{2}=8$

$x=2$

Therefore the length of LM is 6 and length of KN is 10.

Draw perpendicular on KN form L and M.

$KN=KA+AE+EN$

$10=6+2(EN)$ (KA=EN, isosceles trapezoid)

$EN=2$

$KE=KN-EN=10-2=8$

Therefore the length of KE is 8.

Use pythagoras theorem is triangle EKM.

$Hypotenuse^2=base^2+perpendicular^2$

$(KM)^2=(KE)^2+(ME)^2$

$(KM)^2=(8)^2+(3\sqrt{5})^2$

$KM^2=64+9(5)$

$KM=\sqrt{109}$

Therefore the length of side KM is $\sqrt{109}$ units.

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3 years ago

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Solve the system by elimination

dmitriy555 [2]

Answer:

Step-by-step explanation:

Given system is-

-2x+2y+3z=0 -------------(1)

-2x-y+z=-3 ----------------(2)

2x+3y+3z=5 --------------(3)

From (1)-(2) we get

3y+2z=3 ----------------(4)

Again from (2)+(3),

2y+4z=2 ----------------(5)

Now multiplying equation (4) by -2

-6y-4z=-6 --------------(6)

Adding (5) and (6)

-4y=-4

dividing by -4

y=-4 / -4

=1

So, y=1 ,

using y=1 in (5)

2(1)+4z=2

or, 2+4z=2

or, 4z=2-2

or, 4z=0

or, z=0

using y=1 and z=0 in (3)

2x+3(1)+3(0)=5

or, 2x+3+0=5

or, 2x=5-3

or, 2x=2

or, x=2/2

=1

So, the solution is-

(x,y,z)=(1 , 1 , 0)

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3 years ago