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3 years ago

Find the value of x.

Mathematics

1 answer:

mr Goodwill [35]3 years ago

3 0

Given:

A figure of a circle and inscribed quadrilateral JKLM.

$m\angle M=47^\circ,\angle K=(7x+21)^\circ$

To find:

The value of x.

Solution:

The inscribed quadrilateral JKLM in the circle I is a cyclic quadrilateral and the opposite angles of a cyclic quadrilateral are supplementary angles.

Angle M and angle K are opposite angles of a cyclic quadrilateral, it means they are supplementary angles and their sum is 180 degrees.

$m\angle M+m\angle K=180^\circ$

$47^\circ+(7x+21)^\circ=180^\circ$

$(7x+68)^\circ=180^\circ$

$(7x+68)=180$

Isolate the variable x.

$7x=180-68$

$7x=112$

$x=\dfrac{112}{7}$

$x=16$

Therefore, the value of x is 16.

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COLLEGE ALGEBRA!!!!

Savatey [412]

Answer:

$a.A_t=A_o.(\frac{1}{2})^t^/^{t_h}$

$b. 0.010986grams$

$c.109.84 mins$

Step-by-step explanation:

Let $t_1_/_2$ denote the half life of our substance $t$ to be our time, $A_t$ is mass at time t and $A_o$ as initial mass. Our half life equation can be expressed as:

$A_t=A_o.(\frac{1}{2})^t^/^{t_h}$ from which we can obtained mass at any given time.

b. From our equation above,

we can convert our t into mins as $t=4*60=240mins$

We then substitute $t$ value into the equation

$A_2_4_0=45\times(1/2)^2^4^0^/^2^0\\=0.010986\\=0.010986grams$

Hence mass after 4hrs is $0.010986grams$

c. We can set our final mass to $1g$ then substitute in the equation

$A_t=A_o.(\frac{1}{2})^t^/^{t_h}$ . Substitute $A_t=1g$

$1.0=45\times(1/2)^t^/^20$

$1.0=45\times(1/2)^t^/^2^0\\log(1/45)=(t/20)\times log(1/2)\\t=20\times5.491853=109.84 mins$

Hence time until $1g$ mass is 109.83 minutes

5 0

3 years ago

Factor 25b – 14b b(25 – 14) b(25 + 14) -b(25 – 14) -b(25 + 14)

klio [65]

$\huge\text{Hey there!}$

$\large\text{We cannot do much of nothing because we don’t have all the information}\\\large\text{to solve.. so we can just simply do process of elimination until we fully find}\\\large\text{the ANSWER you are trying to look for.}$

$\large\text{Given equation: 25b - 14b}\\\large\text{COMBINE your LIKE TERMS}\\\large\text{\underline{25b - 14b} = 11b}\\\large\text{So, we are looking for an equation that gives us \bf{11b}}$

$\large\text{Option A.}\downarrow\\\large\text{b(25 - 14)}\\\large\text{DISTRIBUTE: b(25) + b(-14)}\\\bullet\large\text{ Reverts to: 25b - 14b (our ORIGINAL EQUATION) so this can be}\\\large\text{equal to 11b}\\\\\large\text{Option A. could POSSIBLY be your ANSWER}$

$\large\text{Option B. }\downarrow\\\large\text{b(25 + 14)}\\\large\text{Distribute}\\\large\text{25(b) + 14(b)}\\\large\text{It DOES NOT revert to the original equation. It gives us: 25b + 14b}\\\large\text{If we COMBINE your LIKE TERMS in that equation. We would get: 39b}\\\large\bf{39b \neq 11b}\\\\\large\text{Option B is NOT your ANSWER}$