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3 years ago

If h(x)=9+x², solve for x when h(x)=13

Mathematics

2 answers:

AlexFokin [52]3 years ago

6 0

Answer:

I hope you got it. I have presented it clearly and stepwise.

vazorg [7]3 years ago

4 0

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \bold{ \sf{x = 2}}}}}}$

Step-by-step explanation:

Given, h(x) = 13

To find : value of x

$\sf{h(x) = 9 + {x}^{2} }$

$\longrightarrow{ \sf{13 = 9 + {x}^{2} }}$

Swap the sides of equation

$\longrightarrow{ \sf{9 + {x}^{2} = 13}}$

Move 9 to right hand side and change it's sign

$\longrightarrow{ \sf{ {x}^{2} = 13 - 9}}$

Subtract 9 from 13

$\longrightarrow{ \sf{ {x}^{2} = 4}}$

$\longrightarrow{ \sf{ \sqrt{ {x}^{2} } = \sqrt{4} }}$

$\longrightarrow{ \sf{x = \: \sqrt{ {2}^{{2}} }}}$

$\longrightarrow{ \sf{x = 2}}$

Hope I helped!

Best regards!

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

The perimeter of a rectangle is about 13.9 cm. If you double the width of the rectangle the perimeter is about 17.8 cm. Find the

fredd [130]

Answer:

5cm by 1.95cm

Step-by-step explanation:

Let the length of the rectangle be x

Let the width be y

Perimeter of a rectangle = 2x + 2y

If the initial perimeter of a rectangle is about 13.9 cm, then;

13.9 = 2x + 2y ..... 1

If when the width is double the perimeter is 17.8cm, then;

17.8 = 2x + 4y ..... 2

Subtract 1 from 2;

13.9 - 17.8 = 2y - 4y

- 3.9 = -2y

y = 3.9/2

y = 1.95 cm

Substitute y = 1.95 into 1 to get x;

From 1;

13.9 = 2x + 2y

13.9 = 2x + 2(1.95)

13.9 = 2x + 3.9

2x = 13.9-3.9

2x = 10

x = 5 cm

Hence the dimension of the smaller triangle is 5cm by 1.95cm

7 0

3 years ago

Mystery Boxes: Breakout Rooms

ollegr [7]

Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

Upper half

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question, the median is:

$Median = 22$ and $N = 14$

To calculate the median, we make use of:

$Median = \frac{N + 1}{2}th$

$Median = \frac{14 + 1}{2}th$

$Median = \frac{15}{2}th$

$Median = 7.5th$

This means that, the median is the average of the 7th and 8th items.

The 7th and 8th items are blanks.

However, from the question; the mode is:

$Mode = 18$

Since the values of the box are in increasing order and the average of 18 and 18 do not equal 22 (i.e. the median), then the 7th item is:

$7th = 18$

The 8th item is calculated as thus:

$Median = \frac{1}{2}(7th + 8th)$

$22= \frac{1}{2}(18 + 8th)$

Multiply through by 2

$44 = 18 + 8th$

$8th = 44 - 18$

$8th = 26$

The updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question.

$Mean = 26$

Mean is calculated as:

$Mean = \frac{\sum x}{n}$

So, we have:

$26= \frac{1 + 2nd + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 12th + 58 + 61}{14}$

Collect like terms

$26= \frac{ 2nd + 12th+1 + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 58 + 61}{14}$

$26= \frac{ 2nd + 12th+304}{14}$

Multiply through by 14

$14 * 26= 2nd + 12th+304$

$364= 2nd + 12th+304$

This gives:

$2nd + 12th = 364 - 304$

$2nd + 12th = 60$

From the updated box,

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We know that:

The 2nd value can only be either 2 or 3

The 12th value can take any of the range 33 to 57

Of these values, the only possible values of 2nd and 12th that give a sum of 60 are:

$2nd = 3$

$12th = 57$

i.e.

$2nd + 12th = 60$

$3 + 57 = 60$

So, the complete box is:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B28%20%7D%20%20%2B%20%20%5Csqrt%7B343%7D%20%20%5Cdiv%202%20%5Csqrt%7B63%20%7D%20"

Nadusha1986 [10]

Answer:

Step-by-step explanation:

sqrt(28): sqrt(4*7)

sqrt(4) = 2;

sqrt28)=2*sqrt(7)

sqrt(343): sqrt(7 * 7 * 7) = 7 * sqrt(7)

Note: the rule is if you have 3 equal primes under the root sign, you leave one, you throw one away, and you put one outside the root sign.

2 sqrt(63) = 2 sqrt(3*3*7) The above rule gets modified to throw 1 three away and take the other one outside the root sign.

2sqrt(63) = 2*3 sqrt(7)

Numerator: 2*sqrt(7) + 7sqrt(7) = 9sqrt(7)

9sqrt(7)

======

6 sqrt(7)

3/2

Note without brackets I cannot be certain that I have interpreted this correctly. The division only apply to sqrt(343) / 2 sqrt(63). If this is so please leave a note.

7 0

3 years ago

Help! What is the answer?

fiasKO [112]

Answer: 22.4

Explanation:

The pieces of the chords multiply together to get the same result. This is from the intersecting chord theorem.

The first chord has pieces of 10 and x, which multiply to 10*x

This is equal to the other product found by multiplying the pieces 14 and 16 (from the other chord)

10*x = 14*16

10x = 224

x = 224/10 ... divide both sides by 10

x = 22.4

5 0

3 years ago