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ipn [44]
2 years ago
11

If x=−4,y=−2,andz=12, solve the following: zx

Mathematics
2 answers:
lara31 [8.8K]2 years ago
7 0

Answer: - 48 (edit: omg i didnt see the minus)

Step-by-step explanation: i mean, only z multiply x

z = 12

x = 4

zx

= z multiply x

= 12 x - 4

= - 48

^_____^

Bezzdna [24]2 years ago
7 0

Answer:

-48

Step-by-step explanation:

zx=12(-4)=-48

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Image attached. Geometry: finding angles
miv72 [106K]

Answer:

28 degrees

Step-by-step explanation:

Because a triangle equals 180, you can subtract 90 and 34 from triangle ABE to leave you with 56 degrees for angle B.

Since CBA is a straigth line with 180 degrees, you can subtract 56 to end up with 124 degrees for angle CBE.

Since line BD bisects angle EBC (splits in half), you can divide the 124 degrees into 62 degrees.

Now that we know angle CBE equals 62 degrees, we can add the 90 degrees and subtract them from the 180 degrees of the triangle.

62+90=152

180-152=28

4 0
3 years ago
What is anwer this qestion and simplife it 8+6²\11 +7×2
Zigmanuir [339]

Answer:

My answer was 18.

Step-by-step explanation:

Step 1:

8+6 times 6 is 44.

Step 2:

Divide 44 by 11 and got 4.

Step 3:

Always multiplication goes first so...

7 times 2 equal 14

Step 4:

14 plus 4

Step 5:

Answer:

18.

4 0
3 years ago
Between 0 degrees Celsius and 30 degrees Celsius, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is giv
Ira Lisetskai [31]

Answer:

T = 3.967 C

Step-by-step explanation:

Density = mass / volume

Use the mass = 1kg and volume as the equation given V, we will come up with the following equation

D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3

   = (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1

Find the first derivative of D with respect to temperature T

dD/dT = \dfrac{70000000\left(291T^2-24298T+91800\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^2}

Let dD/dT = 0 to find the critical value we will get

\dfrac{70000000\left(291T^2-24298T+91800\right)}  = 0

Using formula of quadratic, we get the roots:

T =  79.53 and T = 3.967

Since the temperature is only between 0 and 30, pick T = 3.967

Find 2nd derivative to check whether the equation will have maximum value:

-\dfrac{140000000\left(395178T^4-65993368T^3+3286558821T^2+2886200857800T-121415215620000\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^3}

Substituting the value with T=3.967,

d2D/dT2 = -1.54 x 10^(-8)    a negative value. Hence It is a maximum value

Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of

D = 1/0.001 = 1000 kg/m3

Therefore T = 3.967 C

3 0
3 years ago
Need help with these. 11th grade algebra 2 class. thanks ​
OlgaM077 [116]

15) x=31/10
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3 years ago
A train traveled 399 miles in 7 hours. Find the train's speed.
GrogVix [38]
The train is traveling at 57 miles per hour.
7 0
3 years ago
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