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3 years ago

HELP PLEASE! WILL GIVE BRAINLESIT

Mathematics

1 answer:

Rudik [331]3 years ago

6 0

Answer: Hypotheses ?

Step-by-step explanation:

You might be interested in

1 1\4x2 1\5 ≤ ≥ = 5\6 x 2 1\5

ValentinkaMS [17]

Answer:

≥

Step-by-step explanation:

1 1\4x2 1\5 = 2 3/4

5/6 x 2 1/5 = 1 5/6

7 0

3 years ago

Can the lengths 15 cm, 4 cm, and the square root of 209 cm make a right triangle?

Dimas [21]

Answer: Yes they can

Step-by-step explanation:

Use Pythagoras rule c²=b²+a²

b²=209

c²=15²=225

a²=4²=16

209+16=225

6 0

3 years ago

Are the graphs of the lines in the pair parallel? Explain. y = 4x – 4 24x – 4y = 96

zaharov [31]

No, since the slopes are different

for 1 line
y = 4x - 4

slope = 4

for 2 line
4y = 24x - 96
slope = 24/4 = 6

6 0

3 years ago

Read 2 more answers

(1 point) A rock is thrown into a still pond and causes a circular ripple. If the radius of the ripple is increasing at a rate o

PtichkaEL [24]

Answer:

8pi feet per second

Or, 25.1 feet per second (3 sf)

Step-by-step explanation:

C = 2pi×r

dC/dr = 2pi

dC/dt = dC/dr × dr/dt

= 2pi × 4 = 8pi feet per second

dC/dt = 25.1327412287

8 0

3 years ago

Please answer correctly !!!!!!!!! Will<br> Mark brainliest !!!!!!!!!!!!!

il63 [147K]

Answer:

$x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$

Step-by-step explanation:

$w=-5\left(x-8\right)\left(x+4\right)\\\mathrm{Expand\:}-5\left(x-8\right)\left(x+4\right):\quad -5x^2+20x+160\\w=-5x^2+20x+160\\Switch\:sides\\-5x^2+20x+160=w\\\mathrm{Subtract\:}w\mathrm{\:from\:both\:sides}\\-5x^2+20x+160-w=w-w\\Simplify\\-5x^2+20x+160-w=0\\Solve\:with\:the\:quadratic\:formula\\\mathrm{Quadratic\:Equation\:Formula:}\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-5,\:b=20,\:c=160-w:\quad x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}\\x=\frac{-20+\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20+\sqrt{-20w+3600}}{10}\\x=\frac{-20-\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20-\sqrt{-20w+3600}}{10}\\The\:solutions\:to\:the\:quadratic\:equation\:are\\x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$

6 0

3 years ago

HELP PLEASE! WILL GIVE BRAINLESIT ​

HELP PLEASE! WILL GIVE BRAINLESIT