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3 years ago

7

Someone please answer these 6 questions

2 answers:

Georgia [21]3 years ago

4 0

1)7.28, 69.16 2) 69.16 3)7.98 4)18.2 5)122.08 6)46.75$ for kitten hope this help

Nataly [62]3 years ago

4 0

1- you would save 7.28
you would save 9.16
2-you would pay 177.84
3-it'll cost 20.52 for a canary and a tropical fish
4- the kitten would cost 46.8
5- a puppy would cost 313.92
6-you would save 27.36 on a kitten and a parrot

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Allisa [31]

The first and the last are correct.

3 0

3 years ago

Paula can't quite read the board in her math class. She writes down the equation she reads on the board as 2x - 7 = 23. She corr

Dmitrij [34]

Answer:

6

Step-by-step explanation:

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3 years ago

()) A rectangular piece of paper has a perimeter of 22 inches and an area of 28 square

Natasha2012 [34]

Given:

Perimeter of a rectangular paper = 22 inches.

Area of the rectangular paper = 28 square inches.

To find:

The dimensions of the rectangular paper.

Solution:

Let l be the length and w be the width of the rectangular paper.

Perimeter of a rectangle is:

$P=2(l+w)$

Perimeter of a rectangular paper is 22 inches.

$2(l+w)=22$

$l+w=\dfrac{22}{2}$

$l=11-w$ ...(i)

Area of a rectangle is:

$A=lw$

Area of the rectangular paper is 28 square inches.

$28=lw$

Using (i), we get

$28=(11-w)w$

$28=11w-w^2$

$w^2-11w+28=0$

Splitting the middle term, we get

$w^2-7w-4w+28=0$

$w(w-7)-4(w-7)=0$

$(w-7)(w-4)=0$

Using zero product property, we get

$(w-7)=0\text{ and }(w-4)=0$

$w=7\text{ and }w=4$

If $w=7$ , then by using (i)

$l=11-7$

$l=4$

If $w=4$ , then by using (i)

$l=11-4$

$l=7$

Therefore, the dimensions of the paper are either $7\times 4$ or $4\times 7$ .

5 0

3 years ago

On day 2, Jonathan fills water bottles at a constant rate after 1 day of filling up a few

Burka [1]

Answer:

Step-by-step explanation:

5 0

3 years ago

Let 1 + i be one of the root of z^3 + Az^2 + Bz + 26 = 0, then find the real roots of the equation and determine the value of a

ANEK [815]

Answer:

The real root is -13

The value of A is 11 and B is -24

Step-by-step explanation:

The given equation is $z^3+Az^2+Bz+26=0$ .

If $1+i$ is a root then then the complex conjugate $(1-i)$ is also a root.

Let us now derive the quadratic equation that gave rise to this complex roots.

By the factor theorem, we get;

$(z+(-1-i))(z+(-1+i)=z^2-2z+2$

Now the

Now the fully factored cubic polynomial will be of the form:

$(z+c)(x^2-2x+2)=0$

By comparing constant terms; 2c=26. This means c=13

Hence the required polynomial is $(z+13)(x^2-2x+2)=0$

Therefore the real root is z=-13

We now expand fully to get:

$z^3+11z^2-24z+26=0$

By comparing coefficients; A=11, and B=-24

5 0

4 years ago