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bekas [8.4K]
3 years ago
12

I REALLY NEED HELP!! PLEASE ANSWER!!

Mathematics
1 answer:
luda_lava [24]3 years ago
3 0

Answer:

It won’t load

Step-by-step explanation:

Tell me what it is I can help

You might be interested in
Determine which relation is a function. Question 13 options: a) {(3, 0), (– 2, – 2), (7, – 2), (– 2, 0)} b) c) y = 15x + 2 y = 1
antiseptic1488 [7]

Answer:

x=3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,2(y−d),5

Step-by-step explanation:Solving for x. Want to solve for y or solve for d instead?

1 Simplify  0-20−2  to  -2−2.

3,-2,-27,-2-2,02y=1,5x+2d3,−2,−27,−2−2,02y=1,5x+2d

2 Simplify  -2-2−2−2  to  -4−4.

3,-2,-27,-4,02y=1,5x+2d3,−2,−27,−4,02y=1,5x+2d

3 Subtract 2d2d from both sides.

3-2d,-2-2d,-27-2d,-4-2d,02y-2d=1,5x3−2d,−2−2d,−27−2d,−4−2d,02y−2d=1,5x

4 Divide both sides by 1,51,5.

\frac{3-2d}{1},5,\frac{-2-2d}{1},5,\frac{-27-2d}{1},5,\frac{-4-2d}{1},5,\frac{02y-2d}{1},5=x

​1

​

​3−2d

​​ ,5,

​1

​

​−2−2d

​​ ,5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−4−2d

​​ ,5,

​1

​

​02y−2d

​​ ,5=x

5 Factor out the common term 22.

\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-4-2d}{1},5,\frac{02y-2d}{1},5=x

​1

​

​3−2d

​​ ,5,

​1

​

​−2(1+d)

​​ ,5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−4−2d

​​ ,5,

​1

​

​02y−2d

​​ ,5=x

6 Factor out the common term 22.

\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{02y-2d}{1},5=x

​1

​

​3−2d

​​ ,5,

​1

​

​−2(1+d)

​​ ,5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−2(2+d)

​​ ,5,

​1

​

​02y−2d

​​ ,5=x

7 Factor out the common term 22.

\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x

​1

​

​3−2d

​​ ,5,

​1

​

​−2(1+d)

​​ ,5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−2(2+d)

​​ ,5,

​1

​

​2(y−d)

​​ ,5=x

8 Simplify  \frac{3-2d}{1}

​1

​

​3−2d

​​   to  (3-2d)(3−2d).

3-2d,5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3−2d,5,

​1

​

​−2(1+d)

​​ ,5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−2(2+d)

​​ ,5,

​1

​

​2(y−d)

​​ ,5=x

9 Simplify  \frac{-2(1+d)}{1}

​1

​

​−2(1+d)

​​   to  (-2(1+d))(−2(1+d)).

3-2d,5,-2(1+d),5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3−2d,5,−2(1+d),5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−2(2+d)

​​ ,5,

​1

​

​2(y−d)

​​ ,5=x

10 Simplify  \frac{-27-2d}{1}

​1

​

​−27−2d

​​   to  (-27-2d)(−27−2d).

3-2d,5,-2(1+d),5,-27-2d,5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3−2d,5,−2(1+d),5,−27−2d,5,

​1

​

​−2(2+d)

​​ ,5,

​1

​

​2(y−d)

​​ ,5=x

11 Simplify  \frac{-2(2+d)}{1}

​1

​

​−2(2+d)

​​   to  (-2(2+d))(−2(2+d)).

3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,\frac{2(y-d)}{1},5=x3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,

​1

​

​2(y−d)

​​ ,5=x

12 Simplify  \frac{2(y-d)}{1}

​1

​

​2(y−d)

​​   to  (2(y-d))(2(y−d)).

3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,2(y-d),5=x3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,2(y−d),5=x

13 Switch sides.

x=3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,2(y-d),5x=3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,2(y−d),5

Done

5 0
3 years ago
Para ampliar uma sala quadrada, foram acrescentados 4 metros em seu comprimento e 2 metros em sua largura, conforme a figura. Qu
ollegr [7]

Answer:

b) x^2 + 6x + 8

Step-by-step explanation:

Para ampliar la habitación cuadrada, se agregaron 4 metros de largo y 2 metros de ancho, como se muestra en la figura.

Deje que la longitud y el ancho de la habitación antes de la expansión sean x.

La nueva longitud de la sala es (x + 4).

El nuevo ancho de la sala es (x + 2).

La nueva área de la sala será:

(x + 4)(x + 2)\\x^2 + 2x + 4x + 8\\x^2 + 6x + 8\\\\

Esa es la nueva área de la habitación.

3 0
3 years ago
Write an equation for three-fourth of m is 20​
alisha [4.7K]

Answer:

Since, three fourth of \(t\) is \(15\). Hence, the required equation is \(\frac{3t}{4}=15\)

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Question #8 help me plz
valina [46]

Answer:

UH HOLD ON RIGHT QUICK CAN YOU TELL ME THE QUESTION  

Step-by-step explanation:

4 0
2 years ago
IQ scores based on the​ Stanford-Binet IQ test are normally distributed with a mean of 100 and standard deviation 15. If you wer
lianna [129]

Answer:

95 intervals

Step-by-step explanation:

Given that

population mean = 100

standard deviation = 15

number of interval that involve the mean for 95% confidence interval is calculated as

We know that when we measure the 99 percent confidence interval, 99 outof 100 confidence interval are required to provide the mean population.

similarly

Assuming we measure a confidence interval of 95 percent, then we  should expect 95 out of 100 confidence interval to provide the mean population therefore, answer is 95

7 0
3 years ago
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