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2 years ago

Find the Value of a and b in the triangle below select the two answers fro the list of choices

Mathematics

2 answers:

shutvik [7]2 years ago

7 0

Answer:

Angle a is 42 degrees and Angle b is 138 degrees

Step-by-step explanation:

I show my working in the pic above

The only thing that is weird is that 42 degrees is not an option in in your answers but im certain that angle A is 42 degrees

nataly862011 [7]2 years ago

3 0

For angle a it will be 42degrees and for angle b it will be 138degrees and i am 100%sure

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1. Find the measures of two supplementary angles if the difference between the measures of the two angles is 35

Makovka662 [10]

Answer:

The sum they add up to is 180*.

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2 years ago

A recent survey of

Serga [27]

Answer:

The total number of students in a survey is 300.

Let the number of junior male(JM) be x and the number of senior males(SM) be y.

Let the number of junior female(JF) be p and the number of senior males(SF) be q.

It is given that there are 160 males, 80 junior females, 130 seniors.

Since number of males are 160. So the number of females are,

$300-160=140$

Since number of junior females is 80.

$JF+SF=140\\80+SF=140\\SF=60$

Since number of seniors are 130.

$SM+SF=130\\SM+60=130\\SM=70$

Since number of males is 160.

$JM+SM=160\\JM+70=160\\JM=90$

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3 years ago

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horsena [70]

Answer:

Can u show the whole question

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2 years ago

Will mark the brainliest

Sati [7]

Answer:

B. $60x^8$

Step-by-step explanation:

$4x^2\sqrt{5x^4} \cdot 3\sqrt{5x^8} =$

$= 4x^2 \times 3 \sqrt{5x^4 \times 5x^8}$

$= 12x^2 \sqrt{25x^{4+8}}$

$= 12x^2 \sqrt{25} \sqrt{x^{12}}$

$= 12x^2 \times 5 \times x^6$

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6 0

2 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time t > 0 is $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

$\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t$

where, $\text{R}_i_n$ = concentration of salt in the inflow $\times$ input rate of brine solution

and $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

So, $\text{R}_i_n$ = 4 lb/gal $\times$ 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

= 3 gal/min - 2 gal/min

= 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

= $\frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min}$ = $\frac{2A(t)}{500+t} \text{ lb/min }$

Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;

= $\frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }$

or $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

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2 years ago