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RoseWind [281]
3 years ago
6

If x, y, and z are integers greater than 1, and (327)(510)(z) = (58)(914)(xy), then what is the value of x?

Mathematics
1 answer:
algol [13]3 years ago
5 0

Answer:

<u>Statements (1) and (2) TOGETHER are NOT sufficient.</u>

Explanation:

As in the equation (327)(510)(z) = (58)(914)(xy) there are THREE variables in total i.e. "x", "y" and "z" hence minimum three equations are required to find out values of all variables. Hence,

If the given number of equations is equal to total variable used in any of the equation, values of all the variables can be find out otherwise there can be unlimited number of solutions.

So, value of "x" cannot be determined with the given data.

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Three more than six times a number equals -5. what is the number?
DerKrebs [107]

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-4/3

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first we can name this number x

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Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 15 grams
hram777 [196]

Answer:

X(16)=25.71grams

Step-by-step explanation:

let X(t) denote grams of C formed in  t mins.

For X grams of C we have:

\frac{2}{3}Xg of A and \frac{1}{3}Xg of B

Amounts of A,B remaining at any given time is expressed as:

40-\frac{2}{3}Xg of A and  50-\frac{1}{3}Xg  of B

Rate at which C is formed satisfies:

\frac{dX}{dt} \infty(40-\frac{2}{3}X)(50-\frac{1}{3}X)->\frac{dX}{dt}=k(90-X)\\\therefore \frac{dX}{(90-X)^2}=kdt->\int{\frac{dX}{(90-X)^2}} \, =\int {k} \, dt  \\\therefore \frac{1}{90-X}=kt+c->90-X=\frac{1}{kt+c}\\\\X(t)=90-\frac{1}{kt+c}

Apply the initial condition,X(0)=0 ,to the expression above

0=90-\frac{1}{c} \ \ ->c=\frac{1}{90}\\\therefore\\X(t)=90-\frac{1}{kt+\frac{1}{90}} \ \ ->X(t)=90-\frac{90}{90kt+c}

Now at X(8)=15:

15=90-\frac{90}{90\times 8k+1}  \ ->75=\frac{90}{720k+1}\\k=0.0002778

Substitute  in X(t) to get

X(t)=90-\frac{90}{0.0002778t\times 90+1}\\X(t)=90-\frac{90}{0.25t+1}\\But \ t=16\\\therefore X(t)=90-\frac{90}{0.025\times16+1}\\X(t)=25.71

5 0
3 years ago
In the given figure find the value of x and y?<br>Show you works - Thank you!​
Vaselesa [24]

<u>EXPLANATION</u><u>:</u>

In ∆ ABC , ∠ABC = 40°

  • ∠CAB = x°
  • & ∠ACD = 50°

∠ACD is an exterior angle formed by extending BC to D

We know that

The exterior angle of a triangle formed by extending one side is equal to the sum of the opposite interior angles.

∠ACD = ∠CAB + ∠ABC

⇛50° = x° + 40°

⇛x° = 50°-40°

<h3>⇛x° = 10°</h3>

and

In ∆ ACD , AC = CD

⇛ ∠CDA = ∠CAD

Since the angles opposite to equal sides are equal.

Let ∠CDA = ∠CAD = A°

We know that

The sum of all angles in a triangle is 180°

In ∆ ACD,

∠CDA +∠CAD + ∠ACD = 180°

A°+A°+50° = 180°

⇛2A°+50° = 180°

⇛2A° = 180°-50°

⇛2A° = 130°

⇛A° = 130°/2

⇛A° = 65°

now,

∠CDA = ∠CAD = 65°

∠BAC + ∠CAD+y = 180°

Since angles in the same line

10°+65°+y = 180°

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⇛y = 180°-75°

<h3>⇛y = 105°</h3>

<u>Answer</u><u>:</u> Hence, the value of “x” & “y” will be 10° and 105° respectively.

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