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3 years ago

What fraction plus 7/8 = 1

Mathematics

1 answer:

Bess [88]3 years ago

6 0

Answer:

1/8

Step-by-step explanation:

yes

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If ΔGTR ≅ ΔNAO, then finish the following congruence statements and mark the corresponding congruent sides and the corresponding

telo118 [61]

Answer:

∠A ≅ T

O ≅ ∠R

∠G ≅ N

NO ≅ GR

NA ≅ GT

RT ≅ OA

Step-by-step explanation:

Work shown in picture

3 0

3 years ago

the class marks of a distribution are 37 42 and 47 the class limits corresponding to class mark 42 are

I am Lyosha [343]

Answer:

$39.5,44.5$

Step-by-step explanation:

$Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ first\ interval\ be\ u_1,v_1.\\Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ second\ interval\ be\ u_2,v_2.\\Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ third\ interval\ be\ u_3,v_3.\\Hence,\\As\ the\ class\ marks\ are\ uniform\ with\ an\ a\ difference\ of\ 5, the\\ observation\ is\ continuous\ and\ the\ class\ size\ is\ 5\ too.\\Hence,\\v_1=u_2, v_2=u_3\\$ $Now,\\Lets\ consider\ the\ first\ two\ classes.\\37=\frac{u_1+v_1}{2} \\42=\frac{u_2+v_2}{2}\\By\ adding\ the\ equations:\\79= \frac{u_1+v_1}{2}+\frac{u_2+v_2}{2}\\79=\frac{u_1+v_1+u_2+v_2}{2}\\79=\frac{(u_1+v_2)+(v_1+u_2)}{2}\\79=\frac{(u_1+v_2)}{2}+\frac{(v_1+u_2)}{2}\\Now,\\As\ the\ mid-point\ between\ two\ points\ can\ be\ calculated\ by\\ its\ average:\frac{u+v}{2} \\Hence,\\u_2\ lies\ in\ the\ mid-point\ of\ u_1\ and\ v_2.\\u_2=\frac{(u_1+v_2)}{2}\\$

$Hence,\\By\ substituting\ u_2=\frac{(u_1+v_2)}{2},\\79=u_2+\frac{v_1+u_2}{2}\\As\ v_1=u_2[Proven],\\79=u_2+ \frac{u_2+u_2}{2}\\79=u_2+\frac{2u_2}{2}\\79=2u_2\\Hence,\\u_2=\frac{79}{2}\\u_2=39.5\\Now,\ as\ we\ already\ know\ that\ the\ class\ size=5,\\v_2-u_2=5\\Hence,\\v_2=5+u_2\\Here,\\v_2=5+39.5\\v_2=44.5\\Hence,\\The\ upper\ limit\ of\ the\ second\ interval=44.5\\The\ lower\ limit\ of\ the\ second\ interval=39.5\\$

7 0

3 years ago

Help me with these please!!

Aliun [14]

7x + 5y > = 280 (thats greater then or equal)

8 0

4 years ago

The equation of a circle is x2 + y2 + 6x + 4y + 10 = 1. What is this equation written in its standard form?

Over [174]

You need to complete the square in x and in y,

$x^{2} + y^{2} + 6x + 4y + 10 = 1$

First, group the x-terms and the y-terms separately.

$x^{2} + 6x + y^{2} + 4y + 10 = 1$

Move the 10 to the right side by subtracting 10 from both sides.

$x^{2} + 6x + y^{2} + 4y = -9$

Now complete the square in x and in y.
The constant you need to add to complete each square is the square of half of the coefficient of the x or y term. Make sure to add the constants to both sides of the equation.

$x^{2} + 6x + 9 + y^{2} + 4y + 4= -9 + 9 + 4$

$(x + 3)^{2} + (y + 2)^{2} = 4$

8 0

3 years ago

Used photo math for this, i need to explain how i got these answers. can someone explain it for me

diamong [38]

Answer:

Each equation has a value for the variable.

Plug in that value and use distributive property to solve the rest of the equation.

In other words do the opposite of what is shows

Step-by-step explanation:

5 0

3 years ago