Ask question

Ask question

All categories

3 years ago

14

John is a quarterback. This year, he completed 350350 passes, which is 70\%70% of all the passes he's attempted this year.

1 answer:

Nostrana [21]3 years ago

3 0

X represents the number of all passes for the year.
350 passes =70% of all the passes
350 passes =70% of x
0.70x =350
By solving we find
X= 500 passes

You might be interested in

Plzz help Solve for x x ÷3 3/10 =2 2/5

Svet_ta [14]

Answer:

$\huge\boxed{x=7\dfrac{23}{25}}$

Step-by-step explanation:

$x\div3\dfrac{3}{10}=2\dfrac{2}{5}\\\\\text{convert the mixed number to the impropper fraction}\\\\3\dfrac{3}{10}=\dfrac{3\cdot10+3}{10}=\dfrac{33}{10}\\\\2\dfrac{2}{5}=\dfrac{2\cdot5+2}{5}=\dfrac{12}{5}\\\\x\div\dfrac{33}{10}=\dfrac{12}{5}\\\\x\times\dfrac{10}{33}=\dfrac{12}{5}\qquad\text{multiply both sides by}\ \dfrac{33}{10}\\\\x\times\dfrac{10\!\!\!\!\!\diagup}{33\!\!\!\!\!\diagup}\times\dfrac{33\!\!\!\!\!\diagup}{10\!\!\!\!\!\diagup}=\dfrac{12}{5}\times\dfrac{33}{10}\\\\x=\dfrac{396}{50}$

$x=\dfrac{198}{25}\\\\x=\dfrac{175+23}{25}\\\\x=\dfrac{175}{25}+\dfrac{23}{25}\\\\x=7\dfrac{23}{25}$

3 0

3 years ago

Can someone please help me quick I don’t need help with the graphing part, I just need help with stating the coordinates.

hichkok12 [17]

Answer:1

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Please someone help me. Thank you.

KiRa [710]

Answer:

Step-by-step explanation:

take x degree as reference angle

using tan rule

tan x=opposite/adjacent

tan x=6.1/5.1

tan x=1.19

x= $tan^{-1}$ 1.19

x=49.95

x=50 degree

3 0

3 years ago

Diego solved an equation by multiplying both sides of the equation by 6. Then he checked that 6 is the correct solution by subst

Maslowich

I believe your answer would be c

8 0

3 years ago

Read 2 more answers

A circle has centre (3,0) and radius 5. The line y = 2x + k intersects the circle in two points. Find the set

lara [203]

A circle with center $(3,0)$ and radius $5$ has equation

$(x-3)^2+y^2=25 \iff x^2 + y^2- 6 x = 16$

If we substitute $y=2x+k$ in this equation, we have

$x^2+(2x+k)^2-6x=16 \iff 5x^2+(4k-6)x+k^2-16=0$

This equation has two solutions (i.e. the line intersects the circle in two points) if and only if the determinant is greater than zero:

$\Delta=b^2-4ac=(4k-6)^2-4\cdot 5\cdot (k^2-16)>0$

The expression simplifies to

$-4 (k^2 + 12 k - 89)>0 \iff k^2 + 12 k - 89$

The solutions to the associated equation are

$k^2 + 12 k - 89=0 \iff k=-6\pm 5\sqrt{5}$

So, the parabola is negative between the two solutions:

$-6-5\sqrt{5}$

8 0

3 years ago