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3 years ago

Write an equation of the line that passes through the given point and has the given slope.

Mathematics

1 answer:

kenny6666 [7]3 years ago

4 0

Answer:

2y + 3x + 7 = 0

Step-by-step explanation:

line equation = y - y1 = m (x - x1)

y - (-5) = -3/2 * (x - 1)

y + 5 = -3/2 (x - 1)

2y + 10 = -3x + 3

2y + 3x + 7 = 0

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Lines Line D E and Line A B are parallel. Lines D E and A B are parallel. Angles B and D are alternate interior angles. Which an

Feliz [49]

Answer:

Angles B and D

Step-by-step explanation:

Since Lines D E and Line A B are parallel hence

The angles that will represent alternate interior angles will be

Angles B and D

This is because when two parallel lines are cut through by a transversal, the corresponding angles are congruent .

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4 years ago

Read 2 more answers

10 sqrt 45 please show how you did it

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$10\sqrt{45}=10\sqrt{9\cdot5}=10\sqrt{3^2\cdot5}=\\\\=10\cdot3\sqrt5=\boxed{30\sqrt5}$

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3 years ago

4 cm 1 1 1 1 1 4 cm / 5 cm What is the volume of the solid?

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4 years ago

a set v is given, together with definitions of addition and scalar multiplication. determine which properties of a vector space

agasfer [191]

The properties of a vector space are satisfied Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes aren't legitimate are if $v = x ^ 2 1× v=1^ ×x ^ 2 = 1 #V$

Property three does now no longer follow: Suppose that Property three is legitimate, shall we name $v = a * x ^ 2 +bx +c$ the neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently $0 = O + v = (O x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a$

= 0 If O is the neuter, then it ought to restore x², but $0+ x² = (0x²+0x+zero) + (x²+0x+zero) = 1.$ This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant

have additive inverse

$Let r= v ×2x ^ 2 + v × 1x +v0 , w= w ×2x ^ 2 + w × 1x +w0 . We have that\\v+w= (vO + wO) ^ x^ 2 +(vl^ × wl)^ x+ ( v 2^ × w2)• w+v= (wO + vO) ^x^ 2 +(wl^ × vl)x+ ( w 2^ ×v2)$

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could use z = 1 then $v = x ^ 2 w = x ^ 2 + 1\\(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1$

v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3

Since 3x ^ 2 +1 ne x^ 2 +3. then the associativity rule doesnt hold.

$(1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 3\\1^ × (x^ 2 +x)+2^ × (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )\\(1^ ×2)^ ×(x^ 2 +x)=2^ × (x ^ 2 + x) = 2x + 2\\1^ × (2^ × (x ^ 2 + x) )=1^ × (2x+2)=2x^ 2 +2x( ne2x+2)$

Property f doesnt observe because of the switch of variables. for instance, if v = x ^ 2 1 × v=1^ × x ^ 2 = 1 #V

Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes arent legitimate.

Step-with the aid of using-step explanation:

Note that each sum and scalar multiplication entails in replacing the order from that most important coefficient with the impartial time period earlier than doing the same old sum/scalar multiplication.

Property three does now no longer follow: Suppose that Property three is legitimate, shall we name v = a × x ^ 2 +bx +c the neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently $0 = O + v = (O × x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a$

= zero If O is the neuter, then it ought to restore x², but zero $+ x² = (0x²+0x+zero) + (x²+0x+zero) = 1.$ This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant have additive inverse

Let $r= v × 2x ^ 2 + v × 1x +v0 , w= w2x ^ 2 + w × 1x +w0 . \\We have thatv+w= (vO + wO) ^ x^ 2 +(vl^ wl)^x+ ( v 2^ w2)w+v= (wO + vO) ^ x^ 2 +(wl^ vl)x+ ( w 2^v2)$

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could use $z = 1 then v = x ^ 2 w = x ^ 2 + 1(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3\\Since 3x ^ 2 +1 ne x^ 2 +3.$ then the associativity rule doesnt hold.

Note that each expressions are same because of the distributive rule of actual numbers. Also, you could be aware that his assets holds due to the fact in each instances we 'switch variables twice.

$· (1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 31^ * (x^ 2 +x)+2^ * (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )(1^ * 2)^ * (x^ 2 +x)=2^ * (x ^ 2 + x) = 2x + 21^ * (2^ * (x ^ 2 + x) )=1^ ×* (2x+2)=2x^ 2 +2x( ne2x+2)$

Read more about polynomials :

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8 0

1 year ago

Justin is skeet shooting. The height of the skeet is modeled by the equation h= -5t^2 +32t + 2, where h represents the height in

Rama09 [41]

Answer:

0.05 seconds; 2.575 m off the ground

Step-by-step explanation:

This is a system of equations that you need to solve for the variables t and h.

h = -5t² + 32t + 2

h = 31.5t + 1

Since h will be the same when the bullet and skeet collide, then h is the same in both equations and can be set equal to each other and then you can solve for t.

-5t² + 32t + 2 = 31.5t + 1

-5t² + 32t + 2 - 31.5t - 1= 31.5t + 1 - 31.5t -1 (move everything to the left side)

-5t² + 0.5t + 1 = 0 (combine like terms)

Now we can use the quadratic formula to solve for the time.

t = $\frac{-.5 +/- \sqrt{.25 - 4(-5)(1)} }{2(-5)}$ (substitute)

t = $\frac{-.5 +/- \sqrt{.25 + 20} }{-10}$ (multiply)

t = $\frac{-.5 +/- \sqrt{20.25} }{-10}$ (add)

t = $\frac{-5 +/- 4.5}{-10}$ (square root)

t = -0.5/-10 or -9.5/-10 (combine like terms in each situation)

t = 0.05 or .95 (divide)

Since it will only hit it once, the first time it hits is the amount of time, therefore it takes .05 seconds to hit the skeet.

Next, substitute t into one of the equations to find the height at which it occurs.

h = 31.5(0.05) + 1 (substitute)

h = 1.575 + 1 (multiply)

h = 2.575 (add)

Therefore the skeet will be 2.575 meters off the ground.

4 0

3 years ago