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BlackZzzverrR [31]

3 years ago

8

jaden increased the amount of vegetables she eats each day from 5 servings to 9 servings.by what percentage did jaden increase t

he amount of vegetables she eats?

1 answer:

Stolb23 [73]3 years ago

6 0

$9 - 5 = 4 \\ \dfrac{4}{5} = \dfrac{80}{100} = 80\%$

Answer : 80%

Hope this helps. - M

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Mrs. Hall records the heights of 50 students in a spreadsheet. The mean height is 68 inches. After looking at the data again, sh

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67.3 repeating. You multiply 68 times 50, subtract 82 and 86, them divide by 48. If that doesn't work, try dividing by 50. That should work.

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The sum of the first 5 terms of an AP is 30 and the sum of the four term from T6 to T9 (inclusive) is 69. Find the AP

Ainat [17]

Answer: The AP = 1, ⁷/₂, 6, ¹⁷/₂, 11 ..............

Step-by-step explanation:

From the first statement,

S₅ = ⁵/₂(2a + ( n - 1 )d } = 30

5(2a + 4d )d = 60

10a + 20d = 60

reduce to lowest term to easy calculation by dividing through by 10

a + 2d = 6 -----------------------------------1

second statement

sum of the next 4 terms inclusive

T₉ = ⁹/₂(2a + 8d ) = 69

9(2a + 8d ) = 30 + 69

18a + 72d = 99 x 2

18a + 72d = 198

divide through by 18 to reduce to lowest time

a + 4d = 11 ------------------------------------------2

Now solve the two equation simultaneously to find a and d

a + 2d = 6

a + 4d = 11

-2d = -5

d = ⁵/₂.

Now substitute for d to get a

a + 2(⁵/₂) = 6

a + 5 = 6

a = 6 - 5

a = 1.

Therefore the AP = 1 , ⁷/₂ , 6 , ¹⁷/₂ , 11 , ..............

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3t – 2(t – 1) ≥ 5t – 4(2 + t) please help ill give the crown thingy to the first person who gives me an answer

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Answer:

Infinite many solutions

Meaning there is more than one solution

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Cecil had 6 ice cubes. He put 1 ice cube in each glass. In how many glasses did Cecil put ice cubes?

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Help with geometry hw

tiny-mole [99]

QUESTION 1

Let the third side of the right angle triangle with sides $x,6$ be $l$ .

Then, from the Pythagoras Theorem;

$l^2=x^2+6^2$

$l^2=x^2+36$

Let the hypotenuse of the right angle triangle with sides 2,6 be $m$ .

Then;

$m^2=6^2+2^2$

$m^2=36+4$

$m^2=40$

Using the bigger right angle triangle,

$(x+2)^2=m^2+l^2$

$\Rightarrow (x+2)^2=40+x^2+36$

$\Rightarrow x^2+2x+4=40+x^2+36$

Group similar terms;

$x^2-x^2+2x=40+36-4$

$\Rightarrow 2x=72$

$\Rightarrow x=36$

QUESTION 2

Let the hypotenuse of the triangle with sides (x+2),4 be $k$ .

Then, $k^2=(x+2)^2+4^2$

$\Rightarrow k^2=(x+2)^2+16$

Let the hypotenuse of the right triangle with sides 2,4 be $t$ .

Then; we have $t^2=2^2+4^2$

$t^2=4+16$

$t^2=20$

We apply the Pythagoras Theorem to the bigger right angle triangle to obtain;

$[(x+2)+2]^2=k^2+t^2$

$(x+4)^2=(x+2)^2+16+20$

$x^2+8x+16=x^2+4x+4+16+20$

$x^2-x^2+8x-4x=4+16+20-16$

$4x=24$

$x=6$

QUESTION 3

Let the hypotenuse of the triangle with sides (x+8),10 be $p$ .

Then, $p^2=(x+8)^2+10^2$

$\Rightarrow p^2=(x+8)^2+100$

Let the hypotenuse of the right triangle with sides 5,10 be $q$ .

Then; we have $q^2=5^2+10^2$

$q^2=25+100$

$q^2=125$

We apply the Pythagoras Theorem to the bigger right angle triangle to obtain;

$[(x+8)+5]^2=p^2+q^2$

$(x+13)^2=(x+8)^2+100+125$

$x^2+26x+169=x^2+16x+64+225$

$x^2-x^2+26x-16x=64+225-169$

$10x=120$

$x=12$

QUESTION 4

Let the height of the triangle be H;

Then $H^2+4^2=8^2$

$H^2=8^2-4^2$

$H^2=64-16$

$H^2=48$

Let the hypotenuse of the triangle with sides H,x be r.

Then;

$r^2=H^2+x^2$

This implies that;

$r^2=48+x^2$

We apply Pythagoras Theorem to the bigger triangle to get;

$(4+x)^2=8^2+r^2$

This implies that;

$(4+x)^2=8^2+x^2+48$

$x^2+8x+16=64+x^2+48$

$x^2-x^2+8x=64+48-16$

$8x=96$

$x=12$

QUESTION 5

Let the height of this triangle be c.

Then; $c^2+9^2=12^2$

$c^2+81=144$

$c^2=144-81$

$c^2=63$

Let the hypotenuse of the right triangle with sides x,c be j.

Then;

$j^2=c^2+x^2$

$j^2=63+x^2$

We apply Pythagoras Theorem to the bigger right triangle to obtain;

$(x+9)^2=j^2+12^2$

$(x+9)^2=63+x^2+12^2$

$x^2+18x+81=63+x^2+144$

$x^2-x^2+18x=63+144-81$

$18x=126$

$x=7$

QUESTION 6

Let the height be g.

Then;

$g^2+3^2=x^2$

$g^2=x^2-9$

Let the hypotenuse of the triangle with sides g,24, be b.

Then

$b^2=24^2+g^2$

$b^2=24^2+x^2-9$

$b^2=576+x^2-9$

$b^2=x^2+567$

We apply Pythagaoras Theorem to the bigger right triangle to get;

$x^2+b^2=27^2$

This implies that;

$x^2+x^2+567=27^2$

$x^2+x^2+567=729$

$x^2+x^2=729-567$

$2x^2=162$

$x^2=81$

Take the positive square root of both sides.

$x=\sqrt{81}$

$x=9$

QUESTION 7

Let the hypotenuse of the smaller right triangle be; n.

Then;

$n^2=x^2+2^2$

$n^2=x^2+4$

Let f be the hypotenuse of the right triangle with sides 2,(x+3), be f.

Then;

$f^2=2^2+(x+3)^2$

$f^2=4+(x+3)^2$

We apply Pythagoras Theorem to the bigger right triangle to get;

$(2x+3)^2=f^2+n^2$

$(2x+3)^2=4+(x+3)^2+x^2+4$

$4x^2+12x+9=4+x^2+6x+9+x^2+4$

$4x^2-2x^2+12x-6x=4+9+4-9$

$2x^2+6x-8=0$

$x^2+3x-4=0$

$(x-1)(x+4)=0$

$x=1,x=-4$

We are dealing with length.

$\therefore x=1$

QUESTION 8.

We apply the leg theorem to obtain;

$x(x+5)=6^2$

$x^2+5x=36$

$x^2+5x-36=0$

$(x+9)(x-4)=0$

$x=-9,x=4$

We discard the negative value;

$\therefore x=4$

QUESTION 9;

We apply the leg theorem again;

$10^2=x(x+15)$

$100=x^2+15x$

$x^2+15x-100=0$

Factor;

$(x-5)(x+20)=0$

$x=5,x=-20$

Discard the negative value;

$x=5$

QUESTION 10

According to the leg theorem;

The length of a leg of a right triangle is the geometric mean of the lengths of the hypotenuse and the portion of the hypotenuse adjacent to that leg.

We apply the leg theorem to get;

$8^2=16x$

$64=16x$

$x=4$ units.

QUESTION 11

See attachment

Question 12

See attachment

6 0

4 years ago