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2 years ago

5091.43 to 1 decimal place

Mathematics

1 answer:

Elden [556K]2 years ago

6 0

Answer:

5091.4 if rounding, then idk

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The length of a quarter is 0.955 inch. Round the length of a quarter to the nearest tenth of an inch./8925047/32fb89d7?utm_sourc

m_a_m_a [10]

0.2387528472385828522

7 0

2 years ago

2x (3x + 5)<br>=6x^2+20x

QveST [7]

Answer:

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2 years ago

Read 2 more answers

In the Cash Now lottery game there are 8 finalists who submitted entry tickets on time. From these 8 tickets, three grand prize

alexgriva [62]

Answer:

The number of ways is $\left n} \atop {}} \right. P_r = 336$

Step-by-step explanation:

From the question we are told that

The number of tickets are $n = 8$

The number of finalist are $r =3$

Generally the number of way by which this winners can be drawn and arrange in the order of $1^{st} , \ 2nd , \ 3rd$ is mathematically represented as

$\left n} \atop {}} \right. P_r = \frac{n\ !}{(n-r) !}$

substituting values

$\left n} \atop {}} \right. P_r = \frac{ 8!}{(8-3) !}$

$\left n} \atop {}} \right. P_r = \frac{ 8* 7*6*5*4*3*2*1}{ 5*4*3*2*1}$

$\left n} \atop {}} \right. P_r = 336$

3 0

2 years ago

A plane requires 3300 gallons of fuel and 75 minutes to make a round trip to Prague, and it takes 5700 gallons

Snowcat [4.5K]

Answer:

Step-by-step explanation:

3300P + 5700S > = 94500 (thats greater then or equal to)

this one is based on the fuel

75P + 130S > = 2400 (thats greater then or equal to)

this one is based on the minutes

6 0

3 years ago

EASY BRAINLIEST PLEASE HELP!!

butalik [34]

Answer:

see below

Step-by-step explanation:

<h3>Proposition:</h3>

Let the diagonals AC and BD of the Parallelogram ABCD intercept at E. It is required to prove AE=CE and DE=BE

<h3>Proof:</h3>

1)The lines AD and BC are parallel and AC their transversal therefore,

$\displaystyle \angle DAC = \angle ACB \\ \ \qquad [\text{ alternate angles theorem}]$

2)The lines AB and DC are parallel and BD their transversal therefore,

$\displaystyle \angle BD C= \angle ABD \\ \ \qquad [\text{ alternate angles theorem}]$

3)now in triangle ∆AEB and ∆CED

$\displaystyle \angle EAD=\angle ECB$
$\angle EDA=\angle EBC$
$\displaystyle AD=BC$

therefore,

$\displaystyle \Delta AEB \cong \Delta CED$

hence,

AE=CE
DE=BE

Proven

6 0

2 years ago