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3 years ago

Need help with homework, the answers are there I just need the process.

Mathematics

1 answer:

Natalija [7]3 years ago

5 0

Answer:

bruh its a pdf

Step-by-step explanation:

my sad face tho

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-5 -4 -3 -2 -1 0 1 2 3 4 5 -1 <3

Gelneren [198K]

Answer:

-1

Step-by-step explanation:

.........................................................................................

4 0

2 years ago

Before the coffee shop opens the employees bake 40 cookies which is 25 less than the number of donuts they make does the equatio

Andre45 [30]

Answer:

x-25 = 40

Step-by-step explanation:

Let the number of cookies made = x

The employees bake 40 cookies which is 25 less than the number of donuts they make

Number of Cookies is less than Number of Donuts by 25

Therefore the equation that describes the situation is given as:

x-25 = 40 where x is the number of cookies made.

8 0

3 years ago

Read 2 more answers

-104=-8(k+8) K = ? Solution please if can?

maksim [4K]

Let's solve this question by step by step.

Layout equation.

−104=−8(k+8)

Step 1: Simplify both sides of the equation.

−104=−8(k+8)

−104=(−8)(k)+(−8)(8)(Distribute)

−104=−8k+−64

−104=−8k−64

Step 2: Flip the equation.

−8k−64=−104

Step 3: Add 64 to both sides.

−8k−64+64=−104+64

−8k=−40

Step 4: Divide both sides by -8.

-8k/-8=-40/-8

The answer for this problem is k=5.

5 0

3 years ago

Find the area of quadrilateral ABCD

andreev551 [17]

Answer:

A ≈ 28.5

Step-by-step explanation:

a, b, c

P = a + b + c

Semiperimeter s = $\frac{a+b+c}{2}$

A = $\sqrt{s(s-a)(s-b)(s-c)}$

~~~~~~~~~~~~~~~

$P_{ABC}$ = 4.3 + 2.89 + 6.81 = 14

s = 14 ÷ 2 = 7

$A_{ABC}$ = $\sqrt{7(7-4.3)(7-2.89)(7-6.81)}$ = √14.75901 ≈ 3.84

$P_{BCD}$ = 8.59 + 7.58 + 6.81 = 22.98

s = 22.98 ÷ 2 = 11.49

$A_{BCD}$ = $\sqrt{11.49(11.49-8.59)(11.49-7.58)(11.49-6.81)}$ = √609.7343148 ≈ 24.6928

$A_{ABCD}$ = 3.84 + 24.6928 ≈ 28.5

3 0

2 years ago

Which is the value of this expression when p=-2 and q=-1?

saul85 [17]

Answer:

D. 4

Step-by-step explanation:

$[(p^2) (q^{-3}) ]^{-2}.[(p)^{-3}(q)^5] ^{-2}\\\\=[(p^2) (q^{-3}) \times(p)^{-3}(q)^5 ]^{-2}\\\\=[(p^{2}) \times(p)^{-3} \times (q^{-3}) \times(q)^5 ]^{-2}\\\\=[(p^{2-3}) \times (q^{5-3}) ]^{-2}\\\\=[(p^{-1}) \times (q^{2}) ]^{-2}\\\\=(p^{-1\times (-2)}) \times (q^{2\times (-2) }) \\\\=p^{2}\times q^{-4} \\\\= \frac{p^2}{q^4}\\\\= \frac{(-2)^2}{(-1)^4}\\\\= \frac{4}{1}\\\\= 4$

8 0

3 years ago