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2 years ago

Help with number 4 please

Mathematics

1 answer:

Basile [38]2 years ago

3 0

48 boards

Step-by-step explanation:

length of a board = 5½ in = 11/2

Total length should be covered = 22 ft = 264 in (1ft = 12in)

total boards needed = 264 ÷11/2

= 264 × 2/11

= 24 ×2

= 48 boards

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A particular shade of paint is made by mixing 5 parts red paint with 7 parts blue paint. To make this shade, Shannon mixed 12 qu

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Shannon did not mix the correct shade. Cause the correct ratio of blue and red is 7:5. While Shannon mix 12:8=3:2 which is not equal to 7:5.

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3 years ago

A tennis shop sold twenty $100 racquets at an 8% profit and ten $80 bags at a 15% loss. The profit on the combined transaction w

Sergeeva-Olga [200]

So first let me assume that the Total profit of the racquets are 8 percentage and not individually.

Now to find the profit we can do

20 x 100= 2000

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3 years ago

Read 2 more answers

Lucy has two star systems left to visit on her voyage, but her ship is running low on fuel. The first system, KA-77, is 12001200

-Dominant- [34]

Answer:

1348 light years.

Step-by-step explanation:

Please find the attachment.

Let x represent the distance between KA-7 and KA-11.

We have been given that the first system, KA-7, is 1200 light years away while the second system, KA-11, is 1700 light years away. Lucy sees an angle of 52 degrees between KA-7 and KA-11.

We can see from our attachment that Lucy, KA-7 and KA-11 forms a triangle and we will use law of cosines to solve for x.

$(AB)^2=(AC)^2+(BC)^2-2(AC)(BC)*cos (C)$

Upon substituting our given values in above formula we will get,

$x^2=1200^2+1700^2-2*1200*1700*cos (52^o)$

$x^2=1440000+2890000-4080000*0.615661475$

$x^2=4330000-2511898.81933008$

$x^2=1818101.18066992$

Let us take square root of both sides of our equation.

$x=\sqrt{1818101.18066992}$

$x=1348.3698234\approx 1348$

Therefore, KA-7 and KA-11 are approximately 1348 light years apart.

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3 years ago

algol13

Answer: $ 38.69 ; There are 12 people in the crew

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3 years ago

Determine whether each expression is equivalent to 49^2t – 0.5.

vampirchik [111]

Answer:

None of the expression are equivalent to $49^{(2t - 0.5)}$

Step-by-step explanation:

Given

$49^{(2t - 0.5)}$

Required

Find its equivalents

We start by expanding the given expression

$49^{(2t - 0.5)}$

Expand 49

$(7^2)^{(2t - 0.5)}$

$7^2^{(2t - 0.5)}$

Using laws of indices: $(a^m)^n = a^{mn}$

$7^{(2*2t - 2*0.5)}$

$7^{(4t - 1)}$

This implies that; each of the following options A,B and C must be equivalent to $49^{(2t - 0.5)}$ or alternatively, $7^{(4t - 1)}$

A. $\frac{7^{2t}}{49^{0.5}}$

Using law of indices which states;

$a^{mn} = (a^m)^n$

Applying this law to the numerator; we have

$\frac{(7^{2})^{t}}{49^{0.5}}$

Expand expression in bracket

$\frac{(7 * 7)^{t}}{49^{0.5}}$

$\frac{49^{t}}{49^{0.5}}$

Also; Using law of indices which states;

$\frac{a^{m}}{a^n} = a^{m-n}$

$\frac{49^{t}}{49^{0.5}}$ becomes

$49^{t-0.5}}$

This is not equivalent to $49^{(2t - 0.5)}$

B. $\frac{49^{2t}}{7^{0.5}}$

Expand numerator

$\frac{(7*7)^{2t}}{7^{0.5}}$

$\frac{(7^2)^{2t}}{7^{0.5}}$

Using law of indices which states;

$(a^m)^n = a^{mn}$

Applying this law to the numerator; we have

$\frac{7^{2*2t}}{7^{0.5}}$

$\frac{7^{4t}}{7^{0.5}}$

Also; Using law of indices which states;

$\frac{a^{m}}{a^n} = a^{m-n}$

$\frac{7^{4t}}{7^{0.5}}$ = $7^{4t - 0.5}$

This is also not equivalent to $49^{(2t - 0.5)}$

C. $7^{2t}\ *\ 49^{0.5}$

$7^{2t}\ *\ (7^2)^{0.5}$

$7^{2t}\ *\ 7^{2*0.5}$

$7^{2t}\ *\ 7^{1}$

Using law of indices which states;

$a^m*a^n = a^{m+n}$

$7^{2t+ 1}$

This is also not equivalent to $49^{(2t - 0.5)}$

6 0

3 years ago