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3 years ago

6

Find the slope of each graph. Express the answer in simplest form. Please show graph

2 answers:

Sidana [21]3 years ago

7 0

Answer:

Step-by-step explanation:

slope is -3/2. rise/run. There's already a graph there so I'm assuming no need to show one

OlgaM077 [116]3 years ago

6 0

Answer:

slope: -3/2

y intercept: -1

equation: y = -3/2x-1

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-1.2= y/2.4 -1.7<br> Solve for y

evablogger [386]

${\huge{\boxed{\sf{Question}}$

$-1.2= y\div2.4 -1.7$

Solve for y

${\huge{\boxed{\sf{Answer\:with\:explanation }}}$

let's put our equation first

$-1.2= y\div2.4 -1.7$

Flip the equation

$y\div2.4-1.7=-1.2$

Now follow the rule of $BOMDAS/PEMDAS$

${\huge{\boxed{\sf{BOMDAS }}}$

Brackets

Of multiplication

Division

Addition

Subtraction

${\huge{\boxed{\sf{PEMDAS}}}$

Parenthesis

Exponents

Multiplication

Division

Addition

Subtraction

_________________________

${\huge{\boxed{\sf{Solve\:the\:equation}}}$

_________________________

$y\div2.4-1.7=-1.2$

${\underline{simplify}}$

$0.416667y-1.7=-1.2$

add $1.7$ to both sides

$0.416667y=0.5$

Divide both sides by $0.416667$

$y=1.2$

${\huge{\boxed{\sf{Answer}}}$

$y$ would equal $1.2$

4 0

2 years ago

Solve for each unknown angle measure...

zepelin [54]

Answer:

because l1//l2 => z + 65 = 180

y + 65 = 90

x + 65 = 90

<=> z = 115

y = 25

x = 25

4 0

2 years ago

Read 2 more answers

How many pounds go into 2. 5 tons

daser333 [38]

2.5 tons = 5000 pounds

3 0

2 years ago

Please help me on this problem

Anna71 [15]

Answer:

The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

$a = -4, c = 5$
$a = 1, c = 6$
$a = 2, c = 3$
$a = 3, c = 3$

Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

$-4x^{2} -6x+5 = 0$ for $a = -4 , c=5$

$1x^{2} -6x+6 = 0$ for $a = 1 , c=6$

$2x^{2} -6x+3 = 0$ for $a = 2 , c=3$

$3x^{2} -6x+3 = 0$ for $a = 3 , c=3$

$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

$Discriminant = 116$ which is greater than 0 ∴ a real solution.

for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

7 0

3 years ago

Help me with this :)))))))

lora16 [44]

Answer:

QR= 16

Step-by-step explanation:

HOPE THIS HELPS

3 0

3 years ago