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2 years ago

If 317 + x=609 then, x equals?

Mathematics

2 answers:

Dvinal [7]2 years ago

3 0

If 317+x=609, then x is 292.

609-317=292 therefore, x is 292.

lisov135 [29]2 years ago

3 0

292 is the correct answer.
Please give brainliest and have a nice day!

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The function y = square root sign x is translated using the rule (x, y) → (x – 7, y + 2)to create f(x). What is the domain of f(

Mama L [17]

Answer:

Second option {x | x > 7}

Step-by-step explanation:

We have the function

$y = \sqrt{x}$

We know that the square root of a negative number has no solution in the real ones. Therefore the domain of this function is $x > 0$

When applying the transformation:

$(x, y) \to (x - 7, y + 2)$ we have a translation of the original function in 7 units to the right and 2 units to the top:

$f(x) = \sqrt{x-7} + 2$

In the same way we must guarantee that $(x-7)> 0$

Then $x > 7$ .

Therefore the domain of f(x) is {x | x > 7}

7 0

2 years ago

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In equilateral ∆ABC with side a, the perpendicular to side AB at point B intersects extension of median AM in point P. What is t

Sergeeva-Olga [200]

Answer:

Perimeter = (2 + √3)·a

Step-by-step explanation:

Given: ΔABC is equilateral and AB = a

The diagram is given below :

AM is a median , PB ⊥ AB , PM = b

Now, by using properties of equilateral triangle, median is perpendicular bisector and each angle is of 60°.

We get, ∠AMB = 90°. So, by linear pair ∠AMB + ∠PMB = 180° ⇒ ∠PMB = 90°. Also, ∠ABC = 60° and ∠ABP = 90° (given) So, ∠PBM = 30°

Since, AM is perpendicular bisector of BC. So,

$MB = \frac{a}{2}$

Now in ΔAMB , By using Pythagoras theorem

$AB^{2}=AM^{2}+MB^{2}\\AM^{2}=AB^{2}-MB^{2}\\AM^{2}=a^{2}-(\frac{a}{2})^{2}\\AM=\frac{\sqrt{3}\cdot a}{2}$

Now, in ΔBMP :

$sin\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\\\\sin\thinspace 30^{o}=\frac{\text{MB}}{\text{PB}}\\\\PB=\frac{\text{MB}}{\text{sin 30}}\\\\PB=\frac{\frac{a}{2}}{\frac{1}{2}}\implies PB = a\\\\tan\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Base}}\\\\tan\thinspace 30^{o}=\frac{\text{MB}}{\text{PM}}\\\\PM=\frac{\text{MB}}{\text{tan 30}}\\\\PM=\frac{\frac{a}{2}}{\frac{1}{\sqrt3}}\implies PM=b= \frac{\sqrt{3}\cdot a}{2}$

Perimeter of ABM = AB + PB + PM + AM

$\text{Perimeter = }a+a+b+ \frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a + \frac{\sqrt{3}\cdot a}{2} +\frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a +\sqrt{3}\cdot a\\\\=(2+\sqrt3})\cdot a$

Hence, Perimeter of ΔABP = (2 + √3)·a units

3 0

2 years ago

The owner of a garden center pays $12 for each birdbath. She sells them for 55% more than she paid. Decide whether each expressi

Verdich [7]

Answer:

1 no

2 no

3 no

4 yes

5 no

Step-by-step explanation:

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2 years ago

Erik's disabled sailboat is floating stationary 3 miles east and 2 miles north of Kingston. A ferry leaves Kingston heading towa

sergij07 [2.7K]

Answer:

Step-by-step explanation:

Let's suppose that Ballard is an origin with coordinates (0;0)

Ballard is 8 miles south and 1 mile west of Edmonds ⇒ Edmonds is 1 mile east and 8 miles north of Ballard.

Thus, coordinates of Edmonds (0+1; 0+8) (1;8)

Edmonds is 6 miles due east of Kingston. So, is 6 miles due of Edmonds

Thus, Kingston coordinates (1-6;8) (-5:8)

Sailboat is 3 miles east and 2 miles north of kingston

So, coordinates of sailboat are (-5+3; 8+2) (-2;10)

Ferry leaves Kingston towards Edmonds at mph and Edmonds is 6 miles due east of Kingston.

Initial ferry coordinates as calculated above are equal to Kingston coordinates (-5;8)

After 20 minutes the ferry turns to south.

This distance travelled (20min) is d=9mph*20/60h=3 miles

So, ferry travelled 3 miles toward east in 20 min

Then, coordinates became (-5+3;8+0) (-2;8)

Thus, we have a line connecting two points (-5;8) and (-2;8)

Line connecting them has an equation:

y-8 = (8-8)/(-2+5) * (x+5)

y-8=0

y=8 - This is the equation for the first 20 minutes of travel

Then, ferry turns due south (-2;8) and has a vertiacl line

The equation of verical line is x=a, so the equation will be x=2

b) The sailboat has a radar scope that will detect any object within 3 miles of the sailboat.

region looks like a circular disc with a center (-2;10)

(x+2)^2+(y-10)^2=3^2

(x+2)^2+(y-10)^2 <9 (interior of the circular disc)

(x+2)^2+(y-10)^2 >9 (exterior)

The equation of the line joining the Kingston and Edmonds is y=8

the point of intersection:

(x+2)^2+(8-10)^2=9

(x+2)^2=5

x is approximately 0.24; -4.24

(0.24;8) (-4.24;8) - intersection points

c) The ferry exits the radar during the trid due south long x=-2

The points of intersection of the circle and the line x=-2:

(-2+2)^2+(y-10)^2=9

0+(y-10)^2=0

y is 7.13

(-2;7)

d) Takes south turn at (-2;8), then ferry travels 0.8 miles up to point (-2;7) where it exists the radar zone

Speed is 9mph

The time taken to cover 1 mile = t=1/9hr=1/9 *60=6.667min

So, it exists after 6.67 minutes

e) Ferry enters at (-4.24;8) and takes turn due to south at (-2;8)

The distance travelled is = [-4.24+2]=2.24 miles

The time taken to cover these miles is = t=2.24/9 *60=14.93 min

After turning to south ferry remains in radar for 6.7 min

So, it remains in radar zone for 14.93+6.7=21.63min

8 0

2 years ago

in the figure p is the incenter of isosceles rst what type of triangle is rpt? Thank you in advance .

snow_lady [41]

Answer:

$\Delta \text{RPT is an isosceles triangle}$

Step-by-step explanation:

As in the triangle RST,

$RS=ST\ \ \ \Rightarrow m\angle SRT=m\angle STR$

Because if in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another.

Incenter is obtained by the Intersection of a triangle's three angle bisectors and it is equidistant from the three sides of the triangle.

So, RP and TP are the angle bisectors of ∠R and ∠T respectively.

$\Rightarrow \dfrac{m\angle SRT}{2}=\dfrac{m\angle STR}{2}$

$\Rightarrow m\angle PRT=m\angle PTR$

$\Rightarrow PR=PT$

$\Rightarrow \Delta \text{RPT is an isosceles triangle}$

7 0

2 years ago

Read 2 more answers