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3 years ago

How do you write 1.8 × 10^1 in standard form?

Mathematics

1 answer:

gayaneshka [121]3 years ago

7 0

Answer:that would be 18

Step-by-step explanation:

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How many whole sandwiches can be made with 2 pounds of cheese if each sandwich has 3 ounces of cheese on it?

Sidana [21]

2 pounds =32 ounces divide that by 3 its equals about 10 sandwiches

4 0

3 years ago

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Which table represents points on the graph of h(x) = RootIndex 3 StartRoot negative x + 2 EndRoot?

Alex_Xolod [135]

Answer:

Table 3

Step-by-step explanation:

The third one.

We have the function

$h(x) = \sqrt[3]{-x+2}$

Now we will insert values of x in that definition o h(x) and see if the values we obtain match the corresponding y values in the table:

$h(-6) = \sqrt[3]{-(-6)+2}= \sqrt[3]{6+2}= \sqrt[3]{8} = 2\\h(1) = \sqrt[3]{-1+2}= \sqrt[3]{1}= 1\\h(2) = \sqrt[3]{-2+2}= \sqrt[3]{0}= 0\\h(3) = \sqrt[3]{-3+2}= \sqrt[3]{1}= 1\\h(10) = \sqrt[3]{-10+2}= \sqrt[3]{-8}= -2$

We can see that the values match the table 3, so the table 3 represents points on the graph of h(x)

3 0

2 years ago

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Which graph represents the solution set of the inequality x+ 2 6?

NikAS [45]

Answer:-9-8-7 -10

Step-by-step explanation:

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3 years ago

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The sum of four consecutive even integers is -100. Find the absolute value of the smallest integer.

Talja [164]

Answer:

|–28|=28

Step-by-step explanation:

the four consecutive even integers are –22, –24, –26, –28

the absolute of the smallest is |–28|=28

3 0

2 years ago

Find the center that eliminates the linear terms in the translation of 4x^2 - y^2 + 24x + 4y + 28 = 0.(-3, 2)(-3,- 2)(4, 0)

baherus [9]

Step 1

Given;

$4x^2-y^2+24x+4y+28=0$

Required; To find the center that eliminates the linear terms

Step 2

$\begin{gathered} 4x^2-y^2+24x+4y=-28 \\ 4x^2+24x-y^2+4y=-28 \\ Complete\text{ the square }; \\ 4x^2+24x \\ \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=4 \\ b=24 \\ c=0 \end{gathered}$ $\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\frac{24}{2\times4} \\ d=\frac{24}{8} \\ d=3 \end{gathered}$ $\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{24^2}{4\times4} \\ e=0-\frac{576}{16}=-36 \end{gathered}$

Step 3

Substitute a,d,e into the vertex form

$\begin{gathered} a(x+d)^2+e \\ 4(x+_{}3)^2-36 \end{gathered}$ $\begin{gathered} 4(x+3)^2-36-y^2+4y=-28 \\ 4(x+3)^2-y^2+4y=\text{ -28+36} \\ \\ \end{gathered}$

Step 4

Completing the square for -y²+4y

$\begin{gathered} \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=-1 \\ b=4 \\ c=0 \end{gathered}$ $\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\text{ }\frac{4}{2\times-1} \\ d=\frac{4}{-2} \\ d=-2 \end{gathered}$ $\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{4^2}{4\times(-1)} \\ \\ e=0-\frac{16}{-4} \\ e=4 \end{gathered}$

Step 5

Substitute a,d,e into the vertex form

$\begin{gathered} a(y+d)^2+e \\ =-1(y+(-2))^2+4 \\ =-(y-2)^2+4 \end{gathered}$

Step 6

$\begin{gathered} 4(x+3)^2-y^2+4y=\text{ -28+36} \\ 4(x+3)^2-(y-2)^2+4=-28+36 \\ 4(x+3)^2-(y-2)^2=-28+36-4 \\ 4(x+3)^2-(y-2)^2=4 \\ \frac{4(x+3)^2}{4}-\frac{(y-2)^2}{4}=\frac{4}{4} \\ (x+3)^2-\frac{(y-2)^2}{2^2}=1 \end{gathered}$

Step 7

$\begin{gathered} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \\ \text{This is the }form\text{ of a hyperbola.} \\ \text{From here } \\ a=1 \\ b=2 \\ k=2 \\ h=-3 \end{gathered}$

Hence the answer is (-3,2)

4 0

1 year ago