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hammer [34]
3 years ago
14

Solve the equation for x. -6x 8x = -46

Mathematics
1 answer:
snow_lady [41]3 years ago
6 0
-6x + 8x = -46
2x = -46
x = -46/2 = -23
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Write an equation of the line in slope-intercept form that passes through the point (3, 8) with slope 4?
iren [92.7K]
Slope intercept form:
Y = mx + b
m = slope, b = y intercept
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Plug in the point (3,8)
8 = 4(3) + b
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2 years ago
A scale model of a ramp is a right triangular prism as given in this figure. in the actual ramp, the triangular base has a heigh
kakasveta [241]

The surface area of the actual ramp including the underside is 16.8 ft²

The height of the actual ramp = 1.2 feet  

The height of the model = 6 cm

<h3>What is the surface area of a right triangular prism?</h3>

A right triangular prism is a 3-dimensional triangle with two congruent triangular faces and 3 rectangular faces that connect the triangular.

The surface area of a right triangular prism

= (Perimeter of the base × Length of the prism) + (2 × Base Area)

Scale factor = 1.2 feet / 6 cm = 0.2 ft/cm

Actual base = 16 cm × 0.2 ft/cm = 3.2 ft

convert the units

9 cm = 9 cm × 0.2 ft/cm = 1.8 ft

10 cm = 10 cm × 0.2 ft/cm = 2 ft

The surface area of a right triangular prism

= (Perimeter of the base × Length of the prism) + (2 × Base Area)

Surface area of actual ramp = 2(0.5 × 3.2 ft × 1.2 ft) + 2(2 ft × 1.8 ft) + (1.8 ft × 3.2 ft)

= 16.8 ft²

Hence, the surface area of the actual ramp including the underside is 16.8 ft²

Learn more about the area here:

brainly.com/question/25292087

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What is the area of the right triangle below?
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16 it is as easy as multiplying
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6 and -6 |-6| and -6
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Use the Quadratic Formula to solve x2 + 20x + 98 = 0
Lubov Fominskaja [6]
ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta \ \textless \  0\ then\ no\ solution\\\\if\ \Delta =0\ then\ one\ solution\ x_0=\dfrac{-b}{2a}\\\\if\ \Delta \ \textgreater \  0\ then\ two\ solutions\ x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\-----------------------------

x^2+20x+98=0\\a=1;\ b=20;\ c=98\\\\\Delta=20^2-4\cdot1\cdot98=400-392=8 \ \textgreater \  0\\\sqrt\Delta=\sqrt8=\sqrt{4\cdot2}=\sqrt4\cdot\sqrt2=2\sqrt2\\\\x_1=\dfrac{-20-2\sqrt2}{2\cdot1}=\dfrac{-20-2\sqrt2}{2}=-10-\sqrt2\\\\x_2=\dfrac{-20+2\sqrt2}{2\cdot1}=\dfrac{-20+2\sqrt2}{2}=-10+\sqrt2
3 0
3 years ago
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