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3 years ago

What is the length of the diagonal of a 10 cm by 15 cm rectangle?

Mathematics

1 answer:

wariber [46]3 years ago

6 0

Answer:

18.03 cm

Step-by-step explanation:

The two sides of the rectangle are the legs of a right triangle and the diagonal is the hypotenuse. We can use the Pythagorean theorem

a^2 + b^2 = c^2

10^2 +15^2 = c^2

100 +225 = c^2

325 = c^2

Take the square root of each side

sqrt(325) = sqrt(c^2)

18.02775638 =c

Rounding to the nearest hundredth

18.03 =c

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-8 + 8g = 56 work and answer

Vadim26 [7]

Answer:

g=8

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

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2 years ago

Makayla has $8 to buy tickets at the school fair. each ticket costs $1.50. which inequality best represents how many tickets she

forsale [732]

Answer:

N < 5

Step-by-step explanation:

"Makayla has $8 to buy tickets at the school fair. each ticket costs $1.50" can be written as $8 = ($1.50/ticket)*N.

Dividing both sides by ($1.50/ticket) results in

---------------------- = 5 1/3 tickets

($1.50/ticket)

or N < 5.

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3 years ago

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An architect makes a model of a new house. The model shows a tile patio in the backyard. In the model, each tile has a length o

amid [387]

Answer:

1:8 and 1:64

Step-by-step explanation:

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2 years ago

A jeepney ride cost 11 pesos for the first 4 kilometers and each additional integer kilometers adds 3.50 pesos to the fare, use

Nookie1986 [14]

Answer:

$f(d)=\left\{ \begin{matrix} 11 & \text{for} & 0< d\leq 4 \\-3+3.5d & \text{for}& d > 4 \end$

Step-by-step explanation:

Cost of first 4 kilometers ride = 11 pesos.

So, for the first 4 km, the fair is constant i.e.

for 1 km the fare is 11 pesos,

for 2 km the fair is also 11 pesos,

similarly, for 3 km of 4 km, the fare is 11 pesos.

Hence, for the distance $0\geq d\geq 4$ , the fair function,

$f(d)=11\cdots(i)$

After 4 km, there is an increment of $ 3.50 for each kilometer.

So, the fare function up to 5 kilometers,

$f(d)=11+3.5=14.5$

So, the fare function up to 6 kilometers, i.e for the distance $5$ ,

$f(d)=11+2\times3.5=18$

This can be arranged as the fare function up to 6 km,

$f(d)=11+(6-4)\times3.5=18$

Similarly, the fare function up to $d$ kilometer $(n>4)$ ,

$f(d)=11+(d-4)\times3.5$

$\Rightarrrow f(d)=-3+3.5d\cdots(ii)$

Hence, from equations (i) and (ii),

$f(d)=\left\{ \begin{matrix} 11 & \text{for}\; 0< d \leq 4 \\-3+3.5d & \text{for}\; d > 4 \end$

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3 years ago

A transformation of rectangle LMNO results in rectangle L'M'N'O'.

Helen [10]

Answer:

As you can see Rectangle L M NO and Rectangle L'M'N'O' are such that

1. Size of Rectangle L M NO > Size of Rectangle L'M'N'O'

2. Sides of, means length and breadth of Rectangle L M N O is not in proportion to Rectangle L'M'N'O'. i.e

$\frac{LM}{L'M'}=\frac{8}{6}=\frac{3}{2} \\\\\frac{LO}{L'O'}= \frac{4}{4}=1$

3. Out of four possibilities given :

, Reflection and Rotation is not possible because in all the three cases Pre-image and image are not congruent or similar.

As in,Rectangle LMNO →when length and breadth get reduced by scale of 2 and 0 and Dilation by scale factor of< 1 we get Rectangle L'M'N'O'.Centre of dilation may be other than origin.

So, Stretching of preimage i.e Rectangle L M NO→ means Shrinking has taken place to get image Rectangle L'M'N'O'.As shrinking is not an option.

So, Dilation by a factor less than 1, center of dilation other than origin has taken place.

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2 years ago

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